- #1

ElijahRockers

Gold Member

- 270

- 10

*EDIT: I found my mistake. Theta goes from 90 to 270, not -90 to 90. Wrong side of the y axis. That changes last integration to -2 instead of 2, making final answer -126.*Evaluate the given integral by changing to polar coordinates.

[itex]\int\int_R 3(x+y) dA[/itex]

where R is the region that lies to the left of the y-axis between the circles x

^{2}+ y

^{2}= 1 and x

^{2}+ y

^{2}= 16.

## The Attempt at a Solution

First I converted the region to integrate over in terms of theta and r.

It would be [itex]\frac{-\pi}{2}\leq \theta \frac{\pi}{2}[/itex]

and [itex]1 \leq r \leq 4[/itex]

right? Usually this is the part I have problems with, so this may be wrong.

Next I convert dA which will turn into [itex]rdrd\theta[/itex].

Finally, the integrand 3(x+y) turns into 3r(cos(theta) + sin(theta)), and adding the extra r from dA, altogether getting

[itex]\int_\frac{-\pi}{2}^\frac{\pi}{2}\int_{1}^{4} 3r^{2}(\cos\theta + \sin\theta) drd\theta[/itex].

I got 126, but apparently that is wrong... did I set it up right? here's how I solved

The 3(cos + sin) would be a constant with respect to r. integrating r

^{2}from 1 to 4 gives me 21. 3*21=63, which is constant, so i am integrating (cos + sin) from -90° to 90°. I get 2 for this, and 63 times 2 is 126.

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