Thread: Double Integral Question

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?

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

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

Originally Posted by Danny

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

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

I must have read it wrong. Thanks