Double Integral Question

**DCU** · Aug 18th 2009, 04:37 AM

Will the answer to this be 1/6?
I got limts as

upper x = 1 lower x = sqrt(1-y^2)
upper y = 1 lower y = 0

This right?

**Jester** · Aug 18th 2009, 05:18 AM

Originally Posted by DCU

Will the answer to this be 1/6?
I got limts as

upper x = 1 lower x = sqrt(1-y^2)
upper y = 1 lower y = 0

This right?

I got a different answer. It say to switch to polar so

$<br /> \int_0^{\pi}\int_0^1 (r \cos \theta + r \sin \theta + 1)r dr d \theta = \frac{2}{3} + \frac{\pi}{2}<br />$

**DCU** · Aug 18th 2009, 05:25 AM

Originally Posted by Danny

I got a different answer. It say to switch to polar so

$<br /> \int_0^{\pi}\int_0^1 (r \cos \theta + r \sin \theta + 1)r dr d \theta = \frac{2}{3} + \frac{\pi}{2}<br />$

I must have read it wrong. Thanks

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