http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Fri, 25 Jul 2014 06:51:42 -0500http://calc3.askbot.com/question/176/can-anyone-validate-problem-3/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.Thu, 24 Jul 2014 21:30:49 -0500http://calc3.askbot.com/question/176/can-anyone-validate-problem-3/http://calc3.askbot.com/question/176/can-anyone-validate-problem-3/?answer=178#post-id-178My group got the same answer as you.Fri, 25 Jul 2014 06:51:42 -0500http://calc3.askbot.com/question/176/can-anyone-validate-problem-3/?answer=178#post-id-178