Multivariable Calculus

For a function \(f(x,y)\), the points \((a,b)\) such that \(f_x(a,b) = 0\) and \(f_y(a,b) = 0\) are called critical points.

AND condition: For a critical point \((a,b)\), \(f_x(a,b) = 0\) and \(f_y(a,b) = 0\) should hold at the same time.

1. Find all values of \(x\) such that \(f_x(x,y) = 0\) or \(f_y(x,y) = 0\), say \(x_1, x_2, \dots, x_m\)
2. For \(x_1\)
   • put \(x=x_1\) into the equations \(f_x(x,y) = 0\) and \(f_y(x,y) = 0\), to get \(f_x(x_1,y) = 0\) and \(f_y(x_1,y) = 0\)
   • solve for all values of \(y\) such that \(f_x(x_1,y) = 0\) and \(f_y(x_1,y) = 0\), say \(y_{1},y_{2},\dots,y_{m}\)
   • then all \((x_1, y_{1}), (x_1, y_{2}), \dots, (x_1, y_{m})\) are critical points
3. Repeat Step 2 for each \(x_2, \dots, x_n\), to get all the corresponding critical points.

See Example 4 for illustration.

Find all relative minima, relative maxima or saddle points of \(f(x, y) = 8x^2 + 4xy + y^2 + 48x - 24y - 7\).

First, find all partial derivatives and second-order partial derivatives.

\(f_x(x,y) = 16x + 4y + 48\)

\(f_y(x,y) = 4x + 2y -24\)

\(f_{xx}(x,y) = 16\)

\(f_{yy}(x,y) = 2\)

\(f_{xy}(x,y) = f_{yx}(x,y) = 4\)

Second, find all critical points by letting \(f_x = 0\) and \(f_y = 0\), ie,

\(16x + 4y + 48 = 0\),

\(4x + 2y -24 = 0\).

which means \(x = -12\) and \(y = 36\). Hence \((-12, 36)\) is the only critical point of function \(f\).

Third, calculate the discriminant of \(f\) at the critical point \((-12, 36)\), ie,

\(D = f_{xx}(-12, 36) \cdot f_{yy}(-12, 36) - [f_{xy}(-12, 36)]^2 = 16 \cdot 2 - 4^2 = 16\)

Finally, since \(D > 0\) and \(f_{xx} > 0\), we know \((-12, 36)\) is a relative minimum of \(f\).

Find all relative minima, relative maxima or saddle points of \(f(x, y) = -8x^2 + 8xy + y^2 + 48x - 24y - 7\).

First, find all partial derivatives and second-order partial derivatives.

\(f_x(x,y) = -16x + 8y + 48\)

\(f_y(x,y) = 8x + 2y -24\)

\(f_{xx}(x,y) = -16\)

\(f_{yy}(x,y) = 2\)

\(f_{xy}(x,y) = f_{yx}(x,y) = 8\)

Second, find all critical points by letting \(f_x = 0\) and \(f_y = 0\), ie,

\(-16x + 8y + 48 = 0\),

\(8x + 2y -24 = 0\).

which means \(x = 3\) and \(y = 0\). Hence \((3, 0)\) is the only critical point of function \(f\).

Third, calculate the discriminant of \(f\) at the critical point \((3, 0)\), ie,

\(D = f_{xx}(3, 0) \cdot f_{yy}(3, 0) - [f_{xy}(3, 0)]^2 = -16 \cdot 2 - 8^2 = -96\)

Finally, since \(D < 0\), we know \((3, 0)\) is a saddle point of \(f\).

Find all relative minima, relative maxima or saddle points of \(f(x, y) = x^3 + 8y^3 - 48x - 24y - 7\).

First, find all partial derivatives and second-order partial derivatives.

\(f_x(x,y) = 3x^2 - 48 = 3(x^2 - 16) = 3(x + 4)(x - 4)\)

\(f_y(x,y) = 24y^2 -24 = 24(y^2 - 1) = 24(y + 1)(y - 1)\)

\(f_{xx}(x,y) = 6x\)

\(f_{yy}(x,y) = 48y\)

\(f_{xy}(x,y) = f_{yx}(x,y) = 0\)

Second, find all critical points by letting \(f_x = 0\) and \(f_y = 0\), ie,

\(3(x + 4)(x - 4) = 0\),

\(24(y + 1)(y - 1) = 0\).

We see that \(-4\) and \(4\) are the only two values of \(x\) such that \(f_x = 0\) or \(f_y = 0\).

• When \(x = -4\), we have \(f_x(x,y) = 0\). A critical number should make both \(f_x = 0\) and \(f_y = 0\) hold at the same time. To have \(f_y = 0\), we need \(y = -1\) or \(y = 1\). Therefore, \((x,y) = (-4, -1)\) and \((x,y) = (-4, 1)\) are two critical points.
• When \(x = 4\), we have \(f_x(x,y) = 0\). A critical number should make both \(f_x = 0\) and \(f_y = 0\) hold at the same time. To have \(f_y = 0\), we need \(y = -1\) or \(y = 1\). Therefore, \((x,y) = (4, -1)\) and \((x,y) = (4, 1)\) are two critical points.

Hence, there are 4 critical points in total : \((x,y) = (-4, -1), (-4, 1), (4, -1), (4, 1)\).

Third, calculate the discriminant of \(f\) at each critical point:

• at \((x,y) = (-4,-1)\), \(D = f_{xx}(-4, -1) \cdot f_{yy}(-4, -1) - [f_{xy}(-4, -1)]^2 = 1152 > 0\). Since \(f_{xx}(-4, -1) < 0\), \((x,y) = (-4, -1)\) yields a relative maximum
• at \((x,y) = (-4,1)\), \(D = f_{xx}(-4, 1) \cdot f_{yy}(-4, 1) - [f_{xy}(-4, 1)]^2 = -1152 < 0\), so \((x,y) = (-4, 1)\) yields a saddle point
• at \((x,y) = (4,-1)\), \(D = f_{xx}(4, -1) \cdot f_{yy}(4, -1) - [f_{xy}(4, -1)]^2 = -1152 < 0\), so \((x,y) = (4, -1)\) yields a saddle point
• at \((x,y) = (4,1)\), \(D = f_{xx}(4, 1) \cdot f_{yy}(4, 1) - [f_{xy}(4, 1)]^2 = 1152 > 0\). Since \(f_{xx}(4, 1) > 0\), \((x,y) = (4, 1)\) yields a relative minimum.

Find all relative minima, relative maxima or saddle points of \(f(x, y) = (x^2 - 1)(y^2 - 9)\).

First, find all partial derivatives and second-order partial derivatives.

\(f_x(x,y) = (y^2 - 9) \cdot 2x = 2x(y+3)(y-3)\)

\(f_y(x,y) = (x^2 - 1) \cdot 2y = 2(x+1)(x-1)y\)

\(f_{xx}(x,y) = 2(y^2 - 9)\)

\(f_{yy}(x,y) = 2(x^2 - 1)\)

\(f_{xy}(x,y) = f_{yx}(x,y) = 4xy\)

Second, find all critical points by letting \(f_x = 0\) and \(f_y = 0\), ie,

\(2x(y+3)(y-3) = 0\),

\(2(x+1)(x-1)y = 0\).

We see that \(0, -1\) and \(1\) are the values of \(x\) such that \(f_x = 0\) or \(f_y = 0\).

• When \(x = 0\), we have \(f_x(x,y) = 0\). A critical number should make both \(f_x = 0\) and \(f_y = 0\) hold at the same time. Hence let's see how to make \(f_y = 0\) hold. Putting \(x=0\) into the second equation, we get \(2(x+1)(x-1)y = 2\cdot(0+1)\cdot(0-1)\cdot y = -2y = 0\), which means \(y = 0\). Therefore \((x,y) = (0,0)\) is a critical point.
• When \(x = -1\), we have \(f_y(x,y) = 0\). A critical number should make both \(f_x = 0\) and \(f_y = 0\) hold at the same time. Hence let's see how to make \(f_x = 0\) hold. Putting \(x=-1\) into the first equation, we get \(2\cdot(-1)\cdot(y+3)(y-3) = -2(y+3)(y-3) = 0\), which means \(y = -3\) or \(y = 3\). Therefore \((x,y) = (-1, -3)\) and \((x,y) = (-1, 3)\) are two critical points.
• When \(x = 1\), we have \(f_y(x,y) = 0\). A critical number should make both \(f_x = 0\) and \(f_y = 0\) hold at the same time. Hence let's see how to make \(f_x = 0\) hold. Putting \(x=1\) into the first equation, we get \(2\cdot 1\cdot(y+3)(y-3) = 2(y+3)(y-3) = 0\), which means \(y = -3\) or \(y = 3\). Therefore \((x,y) = (1, -3)\) and \((x,y) = (1, 3)\) are two critical points.

Hence, there are 5 critical points in total : \((x,y) = (0,0), (-1, -3), (-1, 3), (1, -3), (1, 3)\).

Third, calculate the discriminant of \(f\) at each critical point:

• at \((x,y) = (0, 0)\), \(D = f_{xx}(0, 0) \cdot f_{yy}(0, 0) - [f_{xy}(0, 0)]^2 = 36 > 0\). Since \(f_{xx}(0, 0) = -18 < 0\), \((x,y) = (0, 0)\) yields a relative maximum
• at \((x,y) = (-1,-3)\), \(D = f_{xx}(-1, -3) \cdot f_{yy}(-1, -3) - [f_{xy}(-1, -3)]^2 = -144 < 0\), so \((x,y) = (-1, -3)\) yields a saddle point
• at \((x,y) = (-1, 3)\), \(D = f_{xx}(-1, 3) \cdot f_{yy}(-1, 3) - [f_{xy}(-1, 3)]^2 = -144 < 0\), so \((x,y) = (-1, 3)\) yields a saddle point
• at \((x,y) = (1,-3)\), \(D = f_{xx}(1, -3) \cdot f_{yy}(1, -3) - [f_{xy}(1, -3)]^2 = -144 < 0\), so \((x,y) = (1, -3)\) yields a saddle point
• at \((x,y) = (1,3)\), \(D = f_{xx}(1, 3) \cdot f_{yy}(1, 3) - [f_{xy}(1, 3)]^2 = -144 < 0\), so \((x,y) = (1, 3)\) yields a saddle point

This method just helps us find relative extrema, not tells us whether the relative extremum is a relative minimum or a relative maximum.

Find relative extrema of \(f(x,y) = xy\), subjuect to \(x + y = 10\).

1. The constraint can be rewritten as \(g(x,y) = x + y - 10 = 0\).

2. The Lagrange function is

   \(F(x,y,\lambda) = f(x,y) - \lambda \cdot g(x,y)\) \( = xy - \lambda (x + y - 10)\) \( = xy - \lambda x - \lambda y + 10\lambda\).

3. Find partial derivatives

   \(F_x(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial x} = y - \lambda\)

   \(F_y(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial y} = x - \lambda\)

   \(F_{\lambda}(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial \lambda} = -x - y + 10\)

4. Form the system of equations

   System 1:

   \[ F_x(x,y,\lambda) = y - \lambda = 0 \ \ (1) \]

   \[ F_y(x,y,\lambda) = x - \lambda = 0 \ \ (2) \]

   \[ F_{\lambda}(x,y,\lambda) = -x - y + 10 = 0 \ \ (3) \]

5. Let's solve the System 1 in Step 4 by substitution.

   • Represent \(\lambda\) as a function of \(x\) and \(y\) : from \((1)\) we get \(\lambda = y\ \ (1')\).

   • Replace \(\lambda\) with \(y\) in the other two equations, and we get System 2.

     System 2:

     \[ x - y = 0 \ \ (2') \]

     \[ -x - y + 10 = 0 \ \ (3) \]

   Now we just need solve the System 2 which contains only \(x\) and \(y\). Again, we use substitution method to solve it.

   • Represent \(y\) as a function of \(x\) : from \((2')\) we get \(y = x\ \ (2'')\).
   • Replace \(y\) with \(x\) in equation \((3)\), and we get \(- x - x + 10 = 0 \implies x = 5\)
   • By \((2'')\), we have \(y = x = 5\)

   Therefore, \((x,y) = (5,5)\) yields a relative extremum \(f(5,5) = 5 \cdot 5 =25\)

   Remark: You can use any method to solve these equations. Here I just show you how to apply the subtitution method. It may not be the simplest method, but it always works as far as this course is concerned. Using this "stupid" substitution method sometimes has an advantage: at least you will get the answer even if you are not good at manipulating complicated algebraic operations and tricks.

Find relative extrema of \(f(x,y) = x^2 + 2y^2\), subjuect to \(3x + 4y = 5\).

1. The constraint can be rewritten as \(g(x,y) = 3x + 4y - 5 = 0\).

2. The Lagrange function is

   \(F(x,y,\lambda) = f(x,y) - \lambda \cdot g(x,y)\) \(= x^2 + 2y^2 - \lambda (3x + 4y - 5)\) \(= x^2 + 2y^2 - 3 \lambda x - 4 \lambda y + 5 \lambda\).

3. Find partial derivatives

   \(F_x(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial x} = 2x - 3\lambda\)

   \(F_y(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial y} = 4y - 4\lambda\)

   \(F_{\lambda}(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial \lambda} = -3x - 4y + 5\)

4. Form the system of equations

   System 1:

   \[ F_x(x,y,\lambda) = 2x - 3\lambda = 0 \ \ (1) \]

   \[ F_y(x,y,\lambda) = 4y - 4\lambda = 0 \ \ (2) \]

   \[ F_{\lambda}(x,y,\lambda) = -3x - 4y + 5 = 0 \ \ (3) \]

5. Let's solve the System 1 in Step 4 by substitution.

   • Represent \(\lambda\) as a function of \(x\) and \(y\) : from \((1)\) we get \(\lambda = \frac{2x}{3} \ \ (1')\).

   • Replace \(\lambda\) with \(\frac{2x}{3}\) in the other two equations, and we get System 2.

     System 2:

     \[ 4y - 4 \cdot \frac{2x}{3} = 4y - \frac{8x}{3} = 0 \ \ (2') \]

     \[ -3x - 4y + 5 = 0 \ \ (3) \]

   Now we just need solve the System 2 which contains only \(x\) and \(y\). Again, we use substitution method to solve it.

   • Rrepresent \(y\) as a function of \(x\) : from \((2')\) we get \(y = \frac{2x}{3} \ \ (2'')\).
   • Replace \(y\) with \(\frac{2x}{3}\) in equation \((3)\), and we get \(- 3x - 4 \cdot \frac{2x}{3} + 5 = 0 \implies x = \frac{15}{17}\)
   • By \((2'')\), we have \(y = \frac{2}{3} \cdot \frac{15}{17} = \frac{10}{17}\)

   Therefore, \((x,y) = \left( \frac{15}{17},\frac{10}{17} \right)\) yields a relative extremum \(f\left( \frac{15}{17},\frac{10}{17} \right) = \frac{425}{289}\).

   Remark: You can use any method to solve these equations. Here I just show you how to apply the subtitution method. It may not be the simplest method, but it always works as far as this course is concerned. Using this "stupid" substitution method sometimes has an advantage: at least you will get the answer even if you are not good at manipulating complicated algebraic operations and tricks.

Find relative extrema of \(f(x,y) = x^2 y\), subjuect to \(x + 2y = 3\).

1. The constraint can be rewritten as \(g(x,y) = x + 2y - 3 = 0\).

2. The Lagrange function is

   \(F(x,y,\lambda) = f(x,y) - \lambda \cdot g(x,y)\) \(= x^2 y - \lambda (x + 2y - 3)\) \(= x^2 y - \lambda x - 2 \lambda y + 3 \lambda\).

3. Find partial derivatives

   \(F_x(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial x} = 2xy - \lambda\)

   \(F_y(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial y} = x^2 - 2\lambda\)

   \(F_{\lambda}(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial \lambda} = -x - 2y + 3\)

4. Form the system of equations

   System 1:

   \[ F_x(x,y,\lambda) = 2xy - \lambda = 0 \ \ (1) \]

   \[ F_y(x,y,\lambda) = x^2 - 2\lambda = 0 \ \ (2) \]

   \[ F_{\lambda}(x,y,\lambda) = -x - 2y + 3 = 0 \ \ (3) \]

5. Let's solve the System 1 in Step 4 by substitution.

   • Represent \(\lambda\) as a function of \(x\) and \(y\) : from \((1)\) we get \(\lambda = 2xy \ \ (1')\).

   • Replace \(\lambda\) with \(2xy\) in the other two equations, and we get System 2.

     System 2:

     \[ x^2 - 2 \cdot 2xy = x^2 - 4xy = 0 \ \ (2') \]

     \[ -x - 2y + 3 = 0 \ \ (3) \]

   Now we just need solve the System 2 which contains only \(x\) and \(y\). Again, we use substitution method to solve it.

   • Represent \(x\) as a function of \(y\) : from \((3)\) we get \(x = 3-2y \ \ (3')\).

   • Replace \(x\) with \(3-2y\) in equation \((3')\), and we get

     \((3-2y)^2 - 4(3-2y)y = 0\)

     \(\implies 3^2 + (2y)^2 - 2\cdot 3\cdot 2y - (12 - 8y)y = 0\)

     \(\implies 9 + 4y^2 - 12y - 12y + 8y^2 = 0\)

     \(\implies 12y^2 - 24y + 9 = 0\)

     \(\implies 4y^2 - 8y + 3 = 0\)

     \(\implies (2y-1)(2y-3) = 0\)

     \(y = \frac{1}{2}\) or \(y = \frac{3}{2}\)

   • By \((3')\), we have
     • when \(y = \frac{1}{2}\), \(x = 3 - 2 \cdot \frac{1}{2} = 2\)
     • when \(y = \frac{3}{2}\), \(x = 3 - 2 \cdot \frac{3}{2} = 0\)

   Therefore, \((x,y) = \left( 2,\frac{1}{2} \right)\) and \((x,y) = \left( 0,\frac{3}{2} \right)\) yield two relative extrema \(f\left( 2,\frac{1}{2} \right) = 2^2 \cdot \frac{1}{2} = 2\) and \(f\left( 0,\frac{3}{2} \right) = 0^2 \cdot \frac{3}{2} = 0\) respectively.

   Remark: You can use any method to solve these equations. Here I just show you how to apply the subtitution method. It may not be the simplest method, but it always works as far as this course is concerned. Using this "stupid" substitution method sometimes has an advantage: at least you will get the answer even if you are not good at manipulating complicated algebraic operations and tricks.

Find relative extrema of \(f(x,y) = x^{\frac{1}{4}} y^{\frac{3}{4}}\), subjuect to \(x + 2y = 1\).

1. The constraint can be rewritten as \(g(x,y) = x + 2y - 1 = 0\).

2. The Lagrange function is

   \(F(x,y,\lambda) = f(x,y) - \lambda \cdot g(x,y)\) \(= x^{\frac{1}{4}} y^{\frac{3}{4}} - \lambda (x + 2y - 1)\) \(= x^{\frac{1}{4}} y^{\frac{3}{4}} - \lambda x - 2 \lambda y + \lambda\).

3. Find partial derivatives

   \(F_x(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial x} = \frac{1}{4} \cdot x^{\frac{-3}{4}} y^{\frac{3}{4}} - \lambda\)

   \(F_y(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial y} = x^{\frac{1}{4}} \cdot \frac{3}{4} \cdot y^{\frac{-1}{4}} - 2\lambda\)

   \(F_{\lambda}(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial \lambda} = -x - 2y + 1\)

4. Form the system of equations

   System 1:

   \[ F_x(x,y,\lambda) = \frac{1}{4} \cdot x^{\frac{-3}{4}} y^{\frac{3}{4}} - \lambda = 0 \ \ (1) \]

   \[ F_y(x,y,\lambda) = x^{\frac{1}{4}} \cdot \frac{3}{4} \cdot y^{\frac{-1}{4}} - 2\lambda = 0 \ \ (2) \]

   \[ F_{\lambda}(x,y,\lambda) = -x - 2y + 1 = 0 \ \ (3) \]

5. Let's solve the System 1 in Step 4 by substitution.

   • Represent \(\lambda\) as a function of \(x\) and \(y\) : from \((1)\) we get \(\lambda = \frac{1}{4} \cdot x^{\frac{-3}{4}} y^{\frac{3}{4}} \ \ (1')\).

   • Replace \(\lambda\) with \(\frac{1}{4} \cdot x^{\frac{-3}{4}} y^{\frac{3}{4}}\) in the other two equations, and we get System 2.

     System 2:

     \[ x^{\frac{1}{4}} \cdot \frac{3}{4} \cdot y^{\frac{-1}{4}} - 2 \cdot \frac{1}{4} \cdot x^{\frac{-3}{4}} y^{\frac{3}{4}} = 0 \ \ (2') \]

     \[ -x - 2y + 1 = 0 \ \ (3) \]

   Now we just need solve the system 2 which contains only \(x\) and \(y\). Again, we use substitution method to solve it.

   • Before applying the substitution, let's simplify \((2')\) in some way. We see that \((2')\) has fraction powers and even negative fraction powers. Thus if we directly apply the substitution then those fraction powers still exist. We don't like fraction powers or negative powers, so let's eliminate them.
     • we see that the negative power of \(x\) in \((2')\) is \(\frac{-3}{4}\) and the negative power of \(y\) in \((2')\) is \(\frac{-1}{4}\).

     • so we multiply both sides of \((2')\) by \(x^{\frac{3}{4}} y^{\frac{1}{4}}\) in order to offset those negative powers

       \(x^{\frac{1}{4}} \cdot \frac{3}{4} \cdot y^{\frac{-1}{4}} - 2 \cdot \frac{1}{4} \cdot x^{\frac{-3}{4}} y^{\frac{3}{4}} = 0 \ \ (2')\)

       \(\implies x^{\frac{3}{4}} y^{\frac{1}{4}} \cdot \left( x^{\frac{1}{4}} \cdot \frac{3}{4} \cdot y^{\frac{-1}{4}} - 2 \cdot \frac{1}{4} \cdot x^{\frac{-3}{4}} y^{\frac{3}{4}} \right) = x^{\frac{3}{4}} y^{\frac{1}{4}} \cdot 0\)

       \(\implies x^{\frac{3}{4}} y^{\frac{1}{4}} \cdot \left( x^{\frac{1}{4}} \cdot \frac{3}{4} \cdot y^{\frac{-1}{4}} \right) - x^{\frac{3}{4}} y^{\frac{1}{4}} \cdot \left( 2 \cdot \frac{1}{4} \cdot x^{\frac{-3}{4}} y^{\frac{3}{4}} \right) = 0\)

       \(\implies \frac{3}{4} \cdot x^{\frac{3}{4}} x^{\frac{1}{4}} y^{\frac{1}{4}} \cdot y^{\frac{-1}{4}} - 2 \cdot \frac{1}{4} \cdot x^{\frac{3}{4}} \cdot x^{\frac{-3}{4}} \cdot y^{\frac{1}{4}} \cdot y^{\frac{3}{4}} = 0\)

       \(\implies \frac{3}{4} \cdot x - 2 \cdot \frac{1}{4} \cdot y = 0\)

       \(\implies y = \frac{3x}{2} \ \ (2'')\)

       Note here we don't have any fraction power or negative power any more.

     • Now System 2 becomes

       new System 2:

       \[ y = \frac{3x}{2} \ \ (2'') \]

       \[ -x - 2y + 1 = 0 \ \ (3) \]

   • Now let's solve the new System 2 by substitution.
     • \(y\) is already represent as \(\frac{3x}{2}\) in \((2'')\)

     • Replace \(y\) with \(\frac{3x}{2}\) in equation \((3)\), and we get

       \(-x - 2 \cdot \frac{3x}{2} + 1 = 0\)

       \(\implies -x - 3x + 1 = 0\)

       \(\implies x = \frac{1}{4}\)

     • By \((2'')\), we have \(y = \frac{3}{2} \cdot \frac{1}{4} = \frac{3}{8}\)

   Therefore, \((x,y) = \left( \frac{1}{4}, \frac{3}{8} \right)\) yields a relative extremum \(f\left( \frac{1}{4}, \frac{3}{8} \right) = \left( \frac{1}{4} \right) ^ {\frac{1}{4}} \left( \frac{3}{8} \right) ^ {\frac{3}{4}}\).

   Remark: You can use any method to solve these equations. Here I just show you how to apply the subtitution method. It may not be the simplest method, but it always works as far as this course is concerned. Using this "stupid" substitution method sometimes has an advantage: at least you will get the answer even if you are not good at manipulating complicated algebraic operations and tricks.

Find relative extrema of \(f(x,y,z) = xyz\), subjuect to \(2xy + 2xz + yz = 12\).

1. The constraint can be rewritten as \(g(x,y,z) = 2xy + 2xz + yz - 12 = 0\).

2. The Lagrange function is

   \(F(x,y,z,\lambda) = f(x,y,z) - \lambda \cdot g(x,y,z)\) \(= xyz - \lambda (2xy + 2xz + yz - 12)\) \(= xyz - 2\lambda xy - 2\lambda xz - \lambda yz + 12\lambda\)

3. Find partial derivatives

   \(F_x(x,y,z,\lambda) = \frac{\partial F(x,y,z,\lambda)}{\partial x} = yz - 2\lambda y - 2\lambda z\)

   \(F_y(x,y,z,\lambda) = \frac{\partial F(x,y,z,\lambda)}{\partial y} = xz - 2\lambda x - \lambda z\)

   \(F_z(x,y,z,\lambda) = \frac{\partial F(x,y,z,\lambda)}{\partial z} = xy - 2\lambda x - \lambda y\)

   \(F_{\lambda}(x,y,z,\lambda) = \frac{\partial F(x,y,z,\lambda)}{\partial \lambda} = - 2xy - 2xz - yz + 12\)

4. Form the system of equations

   System 1:

   \[ F_x(x,y,z,\lambda) = yz - 2\lambda y - 2\lambda z = 0 \ \ (1) \]

   \[ F_y(x,y,z,\lambda) = xz - 2\lambda x - \lambda z = 0 \ \ (2) \]

   \[ F_z(x,y,z,\lambda) = xy - 2\lambda x - \lambda y = 0 \ \ (3) \]

   \[ F_{\lambda}(x,y,z,\lambda) = - 2xy - 2xz - yz + 12 = 0 \ \ (4) \]

5. NOTE : the hierarchical structure of the following solution is indicated by the bullet symbol \(\bullet\) and indentation.

   Let's solve the System 1 in Step 4 by substitution.

   • Represent \(\lambda\) as a function of \(x, y\) and \(z\) : from \((1)\) we get \(\lambda = \frac{yz}{2y + 2z} \ \ (1')\).

   • Replace \(\lambda\) with \(\frac{yz}{2y + 2z}\) in the other three equations, and we get

     \(F_y(x,y,z,\lambda) = xz - 2\cdot \frac{yz}{2y + 2z} \cdot x - \frac{yz}{2y + 2z} \cdot z\) \(= \frac{xz(2y+2z)}{2y + 2z} - \frac{2xyz}{2y + 2z} - \frac{yz^2}{2y + 2z}\) \(= \frac{2xyz+2xz^2}{2y + 2z} - \frac{2xyz}{2y + 2z} - \frac{yz^2}{2y + 2z}\) \(= \frac{2xyz+2xz^2 - 2xyz - yz^2}{2y + 2z}\) \(= \frac{(2x - y)z^2}{2y + 2z} = 0\)

     \(\implies (2x - y)z^2 = 0 \ \ (2')\)

     \(F_z(x,y,z,\lambda) = xy - 2\cdot \frac{yz}{2y + 2z} \cdot x - \frac{yz}{2y + 2z} \cdot y\) \(= \frac{xy(2y+2z)}{2y + 2z} - \frac{2xyz}{2y + 2z} - \frac{y^2 z}{2y + 2z}\) \(= \frac{2xy^2+2xyz}{2y + 2z} - \frac{2xyz}{2y + 2z} - \frac{y^2 z}{2y + 2z}\) \(= \frac{2xy^2+2xyz - 2xyz - y^2 z}{2y + 2z}\) \(= \frac{(2x - z)y^2}{2y + 2z} = 0\)

     \(\implies (2x - z)y^2 = 0 \ \ (3')\)

     \(F_{\lambda}(x,y,z,\lambda) = - 2xy - 2xz - yz + 12 = 0 \ \ (4)\)

     Putting \((2')\), \((3')\) and \((4)\) together, we have System 2 which contains only \(x, y\) and \(z\) as follows.

     System 2:

     \[ (2x - y)z^2 = 0 \ \ (2') \]

     \[ (2x - z)y^2 = 0 \ \ (3') \]

     \[ F_{\lambda}(x,y,z,\lambda) = - 2xy - 2xz - yz + 12 = 0 \ \ (4) \]

   Now we just need solve the system 2, which contains only \(x, y\) and \(z\). Again, we use substitution method to solve it.

   From \((2')\), we get

   \[ z = 0 \ \ (5) \textbf{ OR } 2x - y = 0 \implies y = 2x \ \ (6) \]

   Note the relationship between \((5)\) and \((6)\) is OR, not AND, so we don't put them simultaneously into equations \((3')\) and \((4)\) in System 2. Instead, at each time, we just use one of them. As a result, we need discuss it case by case.

   • Case 1 : When \(z = 0 \ \ (5)\) holds

     • Replace \(z\) with \(0\) in equations \((3')\) and \((4)\), and we get System 3.1.

       System 3.1:

       \[ 2xy^2 = 0 \ \ (3'') \]

       \[ -2xy + 12 = 0 \ \ (4') \]

     • Now we just need solve the system 3.1 which contains only \(x\) and \(y\). Again, we use substitution method to solve it.
       • From \((3'')\) we get \[x = 0 \ \ (7) \textbf{ OR } y^2 = 0 \implies y = 0 \ \ (8)\]
       • When \(x = 0 \ \ (7)\) holds, replace \(x\) with \(0\) in equation \((4')\), and we get \(12 = 0\), which never holds, so there is no solution of \(y\) corresponding to \(x=0\)
       • When \(y = 0 \ \ (8)\) holds, replace \(y\) with \(0\) in equation \((4')\), and we get \(12 = 0\), which never holds, so there is no solution of \(x\) corresponding to \(y=0\)
       • Therefore, there is not solution to System 3.1 under either \((7)\) or \((8)\).
     • Therefore, there is no solution to \((x,y,z)\) in Case 1.
   • Case 2 : When \(y = 2x \ \ (6)\) holds

     • Replace \(y\) with \(2x\) in equations \((3')\) and \((4)\), and we get System 3.2.

       System 3.2:

       \[ (2x - z) (2x)^2 = 4(2x - z)x^2 = 0 \ \ (3''') \]

       \[ -2x(2x) - 2xz - 2xz + 12 = -4x^2 - 4xz + 12 = 0 \ \ (4'') \]

     • Now we just need solve the system 3.2 which contains only \(x\) and \(z\). Again, we use substitution method to solve it.

       • From \((3''')\), we get

         \[ x^2 = 0 \implies x = 0 \ \ (9) \textbf{ OR } 2x - z = 0 \implies z = 2x \ \ (10) \]

         Note that the relationship between \((9)\) and \((10)\) is OR, not AND, so we don't put them simultaneously into equations \((4'')\) in System 3.2. Instead, at each time, we just use one of them. As a result, we need discuss it case by case.

       • Case 2.1 : When \(x = 0 \ \ (9)\) holds

         Replace \(x\) with \(0\) in equation \((4'')\), and we get \(12 = 0\), which never holds, so there is no solution of \(z\) corresponding to \(x = 0\).

         Therefore, there is no solution to \((x,y,z)\) in Case 2.1.

       • Case 2.2 : When \(z = 2x \ \ (10)\) holds

         • Replace \(z\) with \(2x\) in equation \((4'')\), and we get \(-4x^2 - 4x(2x) + 12 = 0\), so

           \[ x = 1 \ \ (11) \textbf{ OR } x = -1 \ \ (12) \]

         • When \(x = 1 \ \ (11)\) holds, by \((10)\) we have \(z = 2x = 2 \cdot 1 = 2\) and by \((6)\) we have \(y = 2x = 2 \cdot 1 = 2\). Therefore, \((x,y,z) = (1, 2, 2)\) gives a relative extremum.
         • When \(x = -1 \ \ (12)\) holds, by \((10)\) we have \(z = 2x = 2 \cdot (-1) = -2\) and by \((6)\) we have \(y = 2x = 2 \cdot (-1) = -2\). Therefore, \((x,y,z) = (-1, -2, -2)\) gives a relative extremum.
         • Therefore, \((x,y,z) = (1,2,2)\) and \((x,y,z) = (-1,-2,-2)\) are the only two points giving relative extrema in Case 2.2.
     • Also, in whole Case 2, \((x,y,z) = (1,2,2)\) and \((x,y,z) = (-1,-2,-2)\) are the only 2 points giving relative extrema as Case 2.1 has no solution.
   • Finally, for the whole problem, \((x,y,z) = (1,2,2)\) and \((x,y,z) = (-1,-2,-2)\) are the only 2 points giving relative extrema as Case 1 has no solution. The values of the two extrema are \(f(1,2,2) = 1\cdot 2\cdot 2 = 4\) and \(f(-1, -2, -2) = (-1) \cdot (-2) \cdot (-2) = -4\).

   Remark: You can use any method to solve these equations. Here I just show you how to apply the subtitution method. It may not be the simplest method, but it always works as far as this course is concerned. Using this "stupid" substitution method sometimes has an advantage: at least you will get the answer even if you are not good at manipulating complicated algebraic operations and tricks.

This can be derived as follows. First, we have \(\Delta z \approx dz \ \ (1)\)

\(\Delta z = f(x+dx,y+dy) - f(x,y) \ \ (2)\)

\(dz = f_x(x,y) dx + f_y(x,y) dy \ \ (3)\)

Then putting \((2)\) and \((3)\) into \((1)\) we get

\(f(x+dx,y+dy) - f(x,y) \approx f_x(x,y) dx + f_y(x,y) dy\)

\(\implies f(x+dx,y+dy) \approx f(x,y) + f_x(x,y) dx + f_y(x,y) dy\).

Find the total differential of function \(z = x^2 y\) and compute its values when \((x, y) = (2, 3)\) and \((dx, dy) = (\Delta x, \Delta y) = (0.01, 0.02)\).

Find the partial derivatives

\(f_x(x,y) = \frac{\partial f(x,y)}{\partial x} = \frac{\partial x^2 y}{\partial x} = 2xy\)

\(f_x(x,y) = \frac{\partial f(x,y)}{\partial y} = \frac{\partial x^2 y}{\partial x} = x^2\)

Then the total differential is \(dz = f_x(x,y) dx + f_y(x,y) dy = 2xy dx + x^2 dy\)

When \((x,y) = (2,3)\) and \((dx, dy) = (0.01, 0.02)\), we have \(dz = f_x(x,y) dx + f_y(x,y) dy = 2xy dx + x^2 dy\) \(= 2 \cdot 2 \cdot 3 \cdot 0.01 + 2^2 \cdot 0.02\) \(= 0.2\)

Approximate \(2.02^2 \cdot 2.99\).

Let's identify the relevant quantities first:

• the function is \(f(x,y) = x^2 y\)
• we want to approximate \(f(2.02, 2.99)\)
• \((2.02, 2.99)\) is close to \((2, 3)\) and \(f(2,3)\) is easy to compute, so we should approximate \(f(2.02, 2.99)\) by \(f(2,3)\)
• \(dx = 2.02 - 2 = 0.02\) and \(dy = 2.99 - 3 = -0.01\)

Then calculate

• \(f_x(x,y) = \frac{\partial x^2 y}{\partial x} = 2xy\)
• \(f_x(x,y) = \frac{\partial x^2 y}{\partial y} = x^2\)

Therefore, by the formula \(f(x+dx, y+dy) \approx f(x,y) + f_x(x,y) dx + f_y(x,y) dy\), we have

\(f(2.02, 2.99) = f(2+0.02, 3+(-0.01))\)

\(\approx f(2,3) + f_x(2,3) \cdot 0.02 + f_y(2,3) \cdot (-0.01)\)

\(= 2^2 \cdot 3 + 2 \cdot 2 \cdot 3 \cdot 0.02 + 2^2 \cdot (-0.01)\)

\(= 12.2\)

Let \(f(x,y) = x^2 + xy + y^3\). Then

\(\int f(x,y) dx = \int (x^2 + xy + y^3) dx = \frac{x^3}{3} + \frac{yx^2}{2} + y^3 x + C(y)\) where \(C(y)\) is any function of \(y\).

\(\int f(x,y) dy = \int (x^2 + xy + y^3) dy = x^2 y + \frac{xy^2}{2} + \frac{y^4}{4} + C(x)\) where \(C(x)\) is any function of \(x\).

• \(\int f(x,y) dx \neq \int f(x,y) dy\). That is, the integral of \(f\) with respect to \(x\) is not identical to the integral with respect to \(y\). This is the general situation.
• As we did for univariate functions, the indefinite integral, as the "most general antiderivative", is a class of antiderivatives rather than a single one. One antiderivative in this class can be represented as another one plus a constant.
• Here the "constant" means a function that doesn't have the variable with respect to which the integral is defined. That is, it can be a function of the other variables. To find \(\int f(x,y) dx\), we add \(C(y)\) (an arbitrary function of \(y\)) to one specific antiderivative of function \(f(x,y)\) with respect to \(x\) (ie, \(\frac{x^3}{3} + \frac{yx^2}{2} + y^3 x\)), and the result is just the whole family of antiderivatives of \(f(x,y)\) with respect to \(x\). Check it by finding the partial derivative of \(\frac{x^3}{3} + \frac{yx^2}{2} + y^3 x + C(y)\) with respect to \(x\). The similar situation for \(\int f(x,y) dy\), the indefinite integral with respect to \(y\).
• In \(\int f(x,y) dx\), if we use just a constant number \(C\) instead of \(C(y)\) (an arbitrary function of \(y\)), then the result is still a family of antiderivatives of \(f(x,y)\) with respect to \(x\). However, this family is not "the most general" one – not as general as the one using \(C(y)\) – so it is not the "indefinite integral".

Let \(f(x,y) = x^2 + xy + y^3\). Then

\(\int_0^1 f(x,y) dx = \int_0^1 (x^2 + xy + y^3) dx\)

\(= \frac{x^3}{3} + \frac{yx^2}{2} + y^3 x + C(y) \big|_0^1\)

\(= \left( \frac{x^3}{3} + \frac{yx^2}{2} + y^3 x + C(y) \right) \big|_{x=1} - \left( \frac{x^3}{3} + \frac{yx^2}{2} + y^3 x + C(y) \right) \big|_{x=0}\)

\(= \left( \frac{1^3}{3} + \frac{y \cdot 1^2}{2} + y^3 \cdot 1 + C(y) \right) - \left( \frac{0^3}{3} + \frac{y \cdot 0^2}{2} + y^3 \cdot 0 + C(y) \right)\)

\(= \frac{1}{3} + \frac{y}{2} + y^3\)

\(\int_0^1 f(x,y) dy = \int_0^1 (x^2 + xy + y^3) dy\)

\(= x^2 y + \frac{xy^2}{2} + \frac{y^4}{4} + C(x) \big|_0^1\)

\(= \left( x^2 y + \frac{xy^2}{2} + \frac{y^4}{4} + C(x) \right) \big|_{y=1} - \left( x^2 y + \frac{xy^2}{2} + \frac{y^4}{4} + C(x) \right) \big|_{y=0}\)

\(= \left( x^2 \cdot 1 + \frac{x \cdot 1^2}{2} + \frac{1^4}{4} + C(x) \right) - \left( x^2 \cdot 0 + \frac{x \cdot 0^2}{2} + \frac{0^4}{4} + C(x) \right)\)

\(= x^2 + \frac{x}{2} + \frac{1}{4}\)

• \(\int_0^1 f(x,y) dx \neq \int_0^1 f(x,y) dy\). This is the general situation, as the indefinite integrals \(\int f(x,y) dx \neq \int f(x,y) dy\).
• \(\int_0^1 f(x,y) dx\) is a function of only \(y\), and \(\int_0^1 f(x,y) dy\) is a function of only \(x\). This is the general situation. When calculating the definite integral with respect to \(x\), we replace \(x\) in the antiderivative with the lower and upper integral limits, so \(x\) is eliminated and the result \(\int_0^1 x^2 + xy + y^3 dx = \frac{1}{3} + \frac{y}{2} + y^3\) contains only \(y\). It can be imagined that if we further find the definite integral of \(\frac{1}{3} + \frac{y}{2} + y^3\) with respect to \(y\) (ie, find \(\int_a^b \frac{1}{3} + \frac{y}{2} + y^3 dy\)), then \(y\) will be also eliminated and we will get a constant number. We will see this in next section "Double Integral".
• When we evaluate \(\int_0^1 f(x,y) dx\), the definite integral of \(f(x,y)\) with respect to \(x\), \(C(y)\) doesn't play a role because it is eventually cancelled by \(-C(y)\). This is the general case : we don't really need \(C(y)\) for evaluating the definite integral with respect to \(x\), just as we don't need \(C\) for evaluating the univariate definite integral. The similar for the definite integral with respect to \(y\) and \(C(x)\).

For a non-negative \(f(x,y)\), if \(R = \{ (x,y) : a \le x \le b, c \le y \le d \}\), then the volume of the solid under the graph of \(f\) and over the region \(R\), denoted by either \(\iint_R f(x,y) dy dx\) or \(\iint_R f(x,y) dx dy\), equals either \(\int_a^b \left( \int_c^d f(x,y) dy \right) dx\) or \(\int_c^d \left( \int_a^b f(x,y) dx \right) dy\). That is,

\[ V = \iint_R f(x,y) dy dx = \iint_R f(x,y) dx dy \] \[ = \int_a^b \left( \int_c^d f(x,y) dy \right) dx = \int_c^d \left( \int_a^b f(x,y) dx \right) dy \]

For rectangular \(R\), we always have \(\int_a^b \left( \int_c^d f(x,y) dy \right) dx = \int_c^d \left( \int_a^b f(x,y) dx \right) dy\), which means the order of the two layers of integrals can be exchanged.

Find the volume of the solid under the graph of the function \(f(x,y) = 6x^2 y + e^{x+2y}\) and over the region \(R = \{ (x, y) | 3 \le x \le 5, 1 \le y \le 2 \}\).

To calculate the volume, we need find either \(\int_3^5 \left( \int_1^2 6x^2 y + e^{x+2y} dy \right) dx\) or \(\int_1^2 \left( \int_3^5 6x^2 y + e^{x+2y} dx \right) dy\).

Let's find \(\int_3^5 \left( \int_1^2 6x^2 y + e^{x+2y} dy \right) dx\) now. You can find \(\int_1^2 \left( \int_3^5 6x^2 y + e^{x+2y} dx \right) dy\) by yourself using similar steps.

First, find the inner integral \(\int_1^2 6x^2 y + e^{x+2y} dy\).

• Regarding \(x\) as a constant, we get the indefinite integral of \(6x^2 y + e^{x+2y}\) with respect to \(y\)

  \(\int 6x^2 y + e^{x+2y} dy\)

  \(= \int 6x^2 y dy + \int e^{x+2y} dy\)

  \(= 6x^2 \cdot \frac{y^2}{2} + \frac{e^{x+2y}}{2} + C(x)\)

  \(= 3 x^2 y^2 + \frac{e^{x+2y}}{2} + C(x)\)

  Hence, by FTC we get the definite integral \(\int_1^2 6x^2 y + e^{x+2y} dy\)

  \(= \left( 3 x^2 y^2 + \frac{e^{x+2y}}{2} \right) \big|_{y=1}^{y=2}\)

  \(= \left( 3 x^2 \cdot 2^2 + \frac{e^{x + 2 \cdot 2}}{2} \right) - \left( 3 x^2 \cdot 1^2 + \frac{e^{x+2 \cdot 1}}{2} \right)\)

  \(= \left( 12 x^2 + \frac{e^{x + 4}}{2} \right) - \left( 3 x^2 + \frac{e^{x+2}}{2} \right)\)

  \(= 12 x^2 + \frac{e^{x + 4}}{2} - 3 x^2 - \frac{e^{x + 2}}{2}\)

  \(= 9 x^2 + \frac{e^{x + 4}}{2} - \frac{e^{x + 2}}{2}\)

Second, find the outer integral \(\int_3^5 \left( 9 x^2 + \frac{e^{x + 4}}{2} - \frac{e^{x + 2}}{2} \right) dx\).

• The indefinite integral

  \(\int \left( 9 x^2 + \frac{e^{x + 4}}{2} - \frac{e^{x + 2}}{2} \right) dx\)

  \(= \int 9 x^2 dx + \frac{1}{2} \int e^{x + 4} dx - \frac{1}{2} \int e^{x + 2} dx\)

  \(= 3 x^3 + \frac{e^{x + 4}}{2} - \frac{e^{x + 2}}{2} + C\)

• Hence, by FTC we get the definite integral

  \(\int_3^5 \left( 9 x^2 + \frac{e^{x + 4}}{2} - \frac{e^{x + 2}}{2} \right) dx\)

  \(= \left( 3 x^3 + \frac{e^{x + 4}}{2} - \frac{e^{x + 2}}{2} \right) \big|_{x=3}^{x=5}\)

  \(= \left( 3 \cdot 5^3 + \frac{e^{5 + 4}}{2} - \frac{e^{5 + 2}}{2} \right) - \left( 3 \cdot 3^3 + \frac{e^{3 + 4}}{2} - \frac{e^{3 + 2}}{2} \right)\)

  \(= 294 + \frac{e^9 - e^7 - e^7 + e^5}{2}\)

  \(= 294 + \frac{e^9 - 2e^7 + e^5}{2}\)

Therefore, we have \(V = \iint_R f(x,y) dy dx = \int_3^5 \left( \int_1^2 6x^2 y + e^{x+2y} dy \right) dx = \int_3^5 \left( 9 x^2 + \frac{e^{x + 4}}{2} - \frac{e^{x + 2}}{2} \right) dx = 294 + \frac{e^9 - 2e^7 + e^5}{2}\)

• Case 1 : For a non-negative \(f(x,y)\), if the region \(R = \{ (x,y) : a \le x \le b, g_1(x) \le y \le g_2(x) \}\), then the volume of the solid under the graph of \(f\) and over the region \(R\), denoted by \(\iint_R f(x,y) dy dx\), equals \(\int_a^b \left[ \int_{g_1(x)}^{g_2(x)} f(x,y) dy \right] dx\). That is \[ V = \iint_R f(x,y) dy dx = \int_a^b \left[ \int_{g_1(x)}^{g_2(x)} f(x,y) dy \right] dx \]

• Case 2 : For a non-negative \(f(x,y)\), if the region \(R = \{ (x,y) : h_1(y) \le x \le h_2(y), c \le y \le d \}\), then the volume of the solid under the graph of \(f\) and over the region \(R\), denoted by \(\iint_R f(x,y) dx dy\), equals \(\int_c^d \left[ \int_{h_1(y)}^{h_2(y)} f(x,y) dx \right] dy\). That is

  \[ V = \iint_R f(x,y) dx dy = \int_c^d \left[ \int_{h_1(y)}^{h_2(y)} f(x,y) dx \right] dy \]

• For non-rectangular \(R\), \(\int_a^b \left[ \int_{g_1(x)}^{g_2(x)} f(x,y) dy \right] dx \neq \int_{g_1(x)}^{g_2(x)} \left[ \int_a^b f(x,y) dx \right] dy\), and \(\int_c^d \left[ \int_{h_1(y)}^{h_2(y)} f(x,y) dx \right] dy \neq \int_{h_1(y)}^{h_2(y)} \left[ \int_c^d f(x,y) dx \right] dy\), which means the order of the two layers of integrals can NOT be exchanged.
• When the integral limits of \(y\) are functions of \(x\), we should first calculate \(\int_{g_1(x)}^{g_2(x)} f(x,y) dy\), and then calculate the integral of the whole \(\int_{g_1(x)}^{g_2(x)} f(x,y) dy\) with respect to \(x\).
• When the integral limits of \(x\) are functions of \(y\), we should first calculate \(\int_{h_1(y)}^{h_2(y)} f(x,y) dx\), and then calculate the integral of the whole \(\int_{h_1(y)}^{h_2(y)} f(x,y) dx\) with respect to \(y\).
• For short, always let the inside integral be the one with variable integral limits.

Find the volume of the solid under the graph of the function \(f(x,y) = 6x^2 y + e^{x+2y}\) and over the region \(R = \{ (x, y) | y \le x \le 2y, 1 \le y \le 2 \}\).

Note that here the integral limits of \(x\) are functions of \(y\) and the integral limits of \(y\) are constants, so to calculate the volume, we need first do the integration with respect to \(x\) and then do the integration with respect to \(y\). That is, we need find \(\int_1^2 \left( \int_{y}^{2y} 6x^2 y + e^{x+2y} dx \right) dy\).

First, find the inner integral \(\int_{y}^{2y} 6x^2 y + e^{x+2y} dx\).

• Regarding \(y\) as a constant, we get the indefinite integral of \(6x^2 y + e^{x+2y}\) with respect to \(x\)

  \(\int 6x^2 y + e^{x+2y} dx\)

  \(= \int 6x^2 y dx + \int e^{x+2y} dx\)

  \(= 6y \cdot \frac{x^3}{3} + e^{x+2y} + C(y)\)

  \(= 2 x^3 y + e^{x+2y} + C(y)\)

• Hence, by FTC we get the definite integral

  \(\int_{y}^{2y} 6x^2 y + e^{x+2y} dx\)

  \(= ( 2 x^3 y + e^{x+2y} ) \big|_{x = y}^{x = 2y}\)

  \(= ( 2 \cdot (2y)^3 y + e^{2y+2y} ) - ( 2 y^3 y + e^{y+2y} )\)

  \(= ( 16 y^4 + e^{4y} ) - ( 2 y^4 + e^{3y} )\)

  \(= 16 y^4 + e^{4y} - 2 y^4 - e^{3y}\)

  \(= 14 y^4 + e^{4y} - e^{3y}\)

Second, find the outer integral \(\int_1^2 \left( 14 y^4 + e^{4y} - e^{3y} \right) dy\)

• The indefinite integral

  \(\int 14 y^4 + e^{4y} - e^{3y} dy\)

  \(= \int 14 y^4 dy + \int e^{4y} dy - \int e^{3y} dy\)

  \(= \frac{14y^5}{5} + \frac{e^{4y}}{4} - \frac{e^{3y}}{3} + C\)

• Hence by FTC we get the definite integral

  \(\int_1^2 14 y^4 + e^{4y} - e^{3y} dx\)

  \(= \left( \frac{14y^5}{5} + \frac{e^{4y}}{4} - \frac{e^{3y}}{3} \right) \big|_{y=1}^{y=2}\)

  \(= \left( \frac{14 \cdot 2^5}{5} + \frac{e^{4 \cdot 2}}{4} - \frac{e^{3 \cdot 2}}{3} \right) - \left( \frac{14 \cdot 1^5}{5} + \frac{e^{4 \cdot 1}}{4} - \frac{e^{3 \cdot 1}}{3} \right)\)

  \(= \left( \frac{448}{5} + \frac{e^{8}}{4} - \frac{e^{6}}{3} \right) - \left( \frac{14}{5} + \frac{e^{4}}{4} - \frac{e^{3}}{3} \right)\)

  \(= \frac{448}{5} + \frac{e^{8}}{4} - \frac{e^{6}}{3} - \frac{14}{5} - \frac{e^{4}}{4} + \frac{e^{3}}{3}\)

  \(= \frac{434}{5} + \frac{e^{8} - e^{4}}{4} - \frac{e^{3} - e^{6}}{3}\)

Therefore, we have \(V = \int_1^2 \left( \int_{y}^{2y} 6x^2 y + e^{x+2y} dx \right) dy = \int_1^2 \left( 14 y^4 + e^{4y} - e^{3y} \right) dy = \frac{434}{5} + \frac{e^{8} - e^{4}}{4} - \frac{e^{3} - e^{6}}{3}\)