asked 2014-07-30 09:17:18 -0600
I'm having an issue on quiz 3, question 1. What I got was..
$$\int _0 ^{2\pi} \int _0^1 e^{-r^2} $$ $$= \int _0 ^{2\pi} e ^{-r^2} r \delta \theta \bigg |_0 ^1$$ $$= \int _0 ^{2\pi} e^{-1} \delta \theta $$ $$= 2\pi e^{-1} $$
Which I know isn't right, but I'm not sure where exactly I went wrong?
answered 2014-07-30 12:25:57 -0600
My answer was slightly different because when you u-sub you have to change your bounds of integration.
I'll start here:
$$\int_0^{2\pi}\int_0^1e^u\frac{-1}{2}dud\Theta$$
When $r=1, -r^2=-1$ $$= - \pi \int_0^{-1} e^u du$$ $$= -\pi(e^u)\biggr|_0^{-1}$$ $$= -\pi (e^{-1} - e^0)$$ $$= -\pi(\frac{1}{e} -1)$$ $$= \pi - \frac{\pi}{e}$$
answered 2014-07-30 09:49:09 -0600
There should be a u-substitution in this problem. Here is my answer: $$\int_0^{2\pi}\int_0^1 e^{-r^2}rdrd\Theta$$ $$u=-r^2$$ $$du=-2rdr$$ $$\frac{-1}{2}du=rdr$$ $$\int_0^{2\pi}\int_0^1e^u\frac{-1}{2}dud\Theta$$ $$\frac{-1}{2} \pi \int_0^1 e^u du$$ $$-\pi(e^u)\biggr|_0^1$$ $$-\pi e^1 - (-\pi e^0)$$ $$\pi -\pi e$$
Asked: 2014-07-30 09:17:18 -0600
Seen: 12 times
Last updated: Jul 30 '14
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