Question

Let D be the pentagon with vertices in the (x, y)-plane given by

(1, −1),(−1, −1),(−2, 0),(0, 2),(2, 0). Calculate


`\int\limits^._D\int\limits^._. {\\ \sin((x + y)\pi ) dxdy} \,`.

ignore the dots on the integrals

Answer 1

Using the provided order of integration, we can parameterize
`D` by

`D = \left\{(x,y) \mid -1 \le y \le 2 \text{ and } \max(y-2,-y-2) \le x \le \min(-y+2,y+2)\right\}`

where

`\max(y-2,-y-2) = \begin{cases} -y-2 & \text{when } y<0 \\ y-2 & \text{when } y\ge0\end{cases}`

and

`\min(-y+2,y+2) = \begin{cases} y+2 & \text{when } y<0 \\ -y+2 & \text{when } y\ge0\end{cases}`

Then we have two iterated integrals to compute,

`\displaystyle \iint_D \sin((x+y)\pi) \, dA \\\\ ~~~~= \int_(-1)^0 \int_(-y-2)^(y+2) \sin((x+y)\pi) \, dx \, dy + \int_0^2 \int_(y-2)^(-y+2) \sin((x+y)\pi) \, dx\,dy`

Compute the integrals with respect to
`x`.

`\displaystyle \int_(-y-2)^(y+2) \sin((x+y)\pi) \, dx = -\cos((x+y)\pi) \bigg|_(x=-y-2)^(x=y+2) \\\\ ~~~~~~~~ = -\cos((y+2+y)\pi) + \cos((-y-2+y)\pi) \\\\ ~~~~~~~~ = 1 - \cos((y+1)2\pi)`

`\displaystyle \int_(y-2)^(-y+2) \sin((x+y)\pi) \, dx = -1 + \cos((y-1)2\pi)`

Compute the remaining integrals.

`\displaystyle \int_(-1)^0 \left(1 - \cos((y+1)2\pi)\right) \, dy = \left(y - \frac1{2\pi} \sin((y+1)2\pi)\right)\bigg|_(y=-1)^(y=0) \\\\ ~~~~~~~~ = \left(0 - \frac1{2\pi} \sin(2\pi)\right) - \left(-1 - \frac1{2\pi} \sin(0)\right) = 1`

`\displaystyle \int_0^2 \left(-1 + \cos((y-1)2\pi)\right) \, dy = \left(-y + \frac1{2\pi} \sin((y-1)2\pi)\right)\bigg|_(y=0)^(y=2) \\\\ ~~~~~~~~ = \left(-2 + \frac1{2\pi} \sin(2\pi)\right) - \left(0 + \frac1{2\pi} \sin(-2\pi)\right) = -2`

Then the overall integral has a value of

`\displaystyle \iint_D \sin((x+y)\pi) \, dA = 1 + (-2) = \boxed{-1}`