# Evaluate the double integral \int \int_D y dA , where D is the top half of the disk with center...

## Question:

Evaluate the double integral {eq}\int \int_D y \,dA {/eq} , where D is the top half of the disk with center in the origin and radius 5.

## Polar Coordinates:

We can integrate over a region using polar coordinates just like we did in Cartesian coordinates. The only real difference is in the differential area element, which for polar coordinates is {eq}dA = r\ dr\ d\theta {/eq}. Recall also that

{eq}x = r \cos \theta {/eq}

{eq}y = r \sin \theta {/eq}

{eq}r^2 = x^2+y^2 {/eq}

{eq}\theta = \tan^{-1} \frac{y}{x} {/eq}

{eq}dA = r\ dr\ d\theta {/eq}

We are dealing with the top half of a disk, so we want to use polar coordinates here. Then the region is super simple: {eq}r\leq 5 {/eq} when {eq}\theta \in [0,\pi] {/eq}. And then we plug {eq}y {/eq} and {eq}dA {/eq} in polar coordinates into the integral to find

{eq}\begin{align*} \iint_R y\ dA &= \int_0^{\pi} \int_0^5 (r \sin \theta)\ r\ dr\ d\theta \\ &= \int_0^{\pi} \sin \theta\ d \theta \int_0^5 r^2\ dr \\ &= \left [ -\cos \theta \right ]_0^{\pi} \left [ \frac13r^3 \right ]_0^5 \\ &= \frac13(-(-1)+1)(5^3) \\ &= \frac{250}{3} \\ &\approx 83.3333 \end{align*} {/eq}