Double integrals over non-rectangular regions

Background

• Double integrals

What we're building to

• If you wish to perform an integral over a region of the $xy$‍-plane that is not rectangular, you have to express each of the bounds of the inner integral as a function of the outer variable.

  $\begin{array}{r}{\int }_{y_1}^{y_2}\stackrel{\text{Evaluates to some function of }y}{\overbrace{({\int }_{x_1(y)}^{x_2(y)}f(x,y)dx)}}dy\end{array}$‍

  or alternatively,

  $\begin{array}{r}{\int }_{x_1}^{x_2}\stackrel{\text{Evaluates to some function of }x}{\overbrace{({\int }_{y_1(x)}^{y_2(x)}f(x,y)dy)}}dx\end{array}$‍

The trouble with non-rectangular regions

• Find a formula for slices of area using an integral.
• Use a second integral to add those infinitely many slices of area into a volume.

Integrating over a disk

Total volume: $\begin{array}{r}{\int }_{-1}^1(6-2x^2)\sqrt{1-x^2}dx={\displaystyle \frac{11\pi }{4}}\approx 8.6394\end{array}$‍

Slice the other way: Shark fin region

Lower bound: $x=$‍

Upper bound: $x=$‍

Check

Answer

Lower bound: $x=y^2$‍

Upper bound: $x=y+2$‍

These come directly from the definition of the region given above.

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  $\begin{array}{r}{\int }_{y^2}^{y+2}(x+2y)dx\end{array}$‍

  A

  $\begin{array}{r}{\int }_{y^2}^{y+2}(x+2y)dx\end{array}$‍
• (Choice B)  

  $\begin{array}{r}{\int }_0^2(x+2y)dy\end{array}$‍

  B

  $\begin{array}{r}{\int }_0^2(x+2y)dy\end{array}$‍

Check

Answer

The first choice is correct.

$\begin{array}{r}{\int }_{y^2}^{y+2}(x+2y)dx\end{array}$‍

The integration moves horizontally, as indicated by the "$dx$‍". It is also bounded by the values found in the last question, indicated that it stays within the shark fin region.

Area of constant-$y$‍-value slice:

Check

Answer

$\begin{array}{rl}{\int }_{y^2}^{y+2}(x+2y)dx& ={({\displaystyle \frac{x^2}{2}}+2yx)}_{x=y^2}^{x=y+2}\\ \\ & =({\displaystyle \frac{(y+2{)}^2}{2}}+2y(y+2))-({\displaystyle \frac{(y^2{)}^2}{2}}+2y(y^2))\\ \\ & =\underset{\text{Factor out }1/2}{\underbrace{{\displaystyle \frac{y^2+4y+4}{2}}+2y^2+4y-{\displaystyle \frac{y^4}{2}}-2y^3}}\\ \\ & ={\displaystyle \frac{1}{2}}(y^2+4y+4+4y^2+8y-y^4-4y^3)\\ \\ & ={\displaystyle \frac{1}{2}}(-y^4-4y^3+5y^2+12y+4)\\ \end{array}$‍

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  $\begin{array}{r}{\int }_0^4\dots dy\end{array}$‍

  A

  $\begin{array}{r}{\int }_0^4\dots dy\end{array}$‍
• (Choice B)  

  $\begin{array}{r}{\int }_0^2\dots dy\end{array}$‍

  B

  $\begin{array}{r}{\int }_0^2\dots dy\end{array}$‍

Check

Answer

Looking at the picture of our shark fin region, $y$‍ varies from $0$‍ to $2$‍

Therefore, the second choice is correct. The integral giving our desired volume looks like this:

$\begin{array}{r}{\int }_0^2{\displaystyle \frac{1}{2}}(-y^4-4y^3+5y^2+12y+4)dy\end{array}$‍

Volume:

Check

Answer

$\begin{array}{rl}& {\int }_0^2{\displaystyle \frac{1}{2}}(-y^4-4y^3+5y^2+12y+4)dy\\ \\ & ={\displaystyle \frac{1}{2}}{(-{\displaystyle \frac{y^5}{5}}-y^4+5{\displaystyle \frac{y^3}{3}}+6y^2+4y)}_0^2\\ \\ & ={\displaystyle \frac{1}{2}}(-{\displaystyle \frac{(2{)}^5}{5}}-(2{)}^4+5{\displaystyle \frac{(2{)}^3}{3}}+6(2{)}^2+4(2))\\ & -{\displaystyle \frac{1}{2}}(-{\displaystyle \frac{(0{)}^5}{5}}-(0{)}^4+5{\displaystyle \frac{(0{)}^3}{3}}+6(0{)}^2+4(0))\\ \\ & ={\displaystyle \frac{1}{2}}(-{\displaystyle \frac{32}{5}}-16+5{\displaystyle \frac{8}{3}}+24+8)\\ \\ & =\dots (\text{calculator})\dots \\ \\ & ={\displaystyle \frac{172}{15}}\end{array}$‍

Summary

• Start by cutting your region along slices that correspond with holding one of the variables constant. For example, holding $x$‍ at some constant value will give a vertical stripe of your region.
• Find how to express the bounds of these stripes as a function of the other variable. For example, the top and bottom of a vertical stripe would be expressed as some function of $x$‍.
• When you set up your double integral, the inner integral will correspond to integrating along one of these stripes, and each of its bounds will be a function of the outer variable. If the inner integral corresponds to constant-$x$‍-values, the double integral as a whole might look like this:
  $\begin{array}{r}{\int }_{x_1}^{x_2}\stackrel{\text{Evaluates to some function of }x}{\overbrace{({\int }_{y_1(x)}^{y_2(x)}f(x,y)dy)}}dx\end{array}$‍
  Alternatively, if you started with horizontal constant-$y$‍-value slices, the double integral might look like this:
  $\begin{array}{r}{\int }_{y_1}^{y_2}\stackrel{\text{Evaluates to some function of }y}{\overbrace{({\int }_{x_1(y)}^{x_2(y)}f(x,y)dx)}}dy\end{array}$‍