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?
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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
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