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Can Anyone Validate Problem #3

asked 2014-07-24 21:30:49 -0600

Gear Junky gravatar image

So for problem #3 on the In-Class worksheet I worked it through, but didn't have enough time to validate my answer with the rest of my group. I approached it as: $$\int_0^{2\pi} \int_0^{\sqrt{\pi/2}} \int_0^{\cos(r^2)} Z \delta z r \delta r \delta \theta$$ $$=2\pi \int_0^{\sqrt{\pi/2}} \int_0^{\cos(r^2)} Z \delta z r \delta r$$ $$=2\pi \int_0^{\sqrt{\pi/2}} \frac{\cos^2(r^2)}{2} r \delta r$$ And by the Double angle formula I got: $$=\frac{\pi}{2}\int_0^{\sqrt{\pi/2}}r(1+\cos(2r^2)\delta r$$ Using U-sub: $$u=2{r^2}$$ $$\delta u=4r \delta r$$ So I ended up with: $$=\frac{\pi}{8} \int_0^\pi 1+\cos(u) \delta u$$ $$=\frac{\pi}{8}(u+\sin(u)) |_0^\pi$$ Getting an answer of: $$\frac{\pi^2}{8}$$ If anyone got a different answer or can see an error in my math I would greatly appreciate your input.

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answered 2014-07-25 06:51:42 -0600

Anonymous gravatar image

My group got the same answer as you.

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Asked: 2014-07-24 21:30:49 -0600

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Last updated: Jul 25 '14