Question # 01: Evaluate the iterated integral by converting to polar coordinates The region of integration is bounded by Question # 02: Evaluate the triple integral
Last edited by sajjad002; May 26th 2009 at 01:49 AM.
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The first is just $\displaystyle \int_{-\pi/2}^{\pi/2}\int_0^2 re^{r^2}dr d\theta =\pi\int_0^2 re^{r^2}dr$. WHERE I think you have $\displaystyle x^2+y^2$ in the exponent of e, but IT's hard to read. Now integrate with $\displaystyle u=r^2$.
Originally Posted by matheagle the first is just $\displaystyle \int\int re^{r^2}dr d\theta$ WHERE I think you have $\displaystyle x^2+y^2$ in the exponent of e, but IT's hard to read. $\displaystyle \int_{-0}^{2}\int_{-(4-x^2)^{1/2}}^{(4-x^2)^{1/2}} e^{-x^2-y^2}dy dx$
The second one is $\displaystyle \int_2^3dx\int_{-1}^1\int_0^1 (\sqrt{y}-3z^2)dydz=\int_{-1}^1\int_0^1(\sqrt{y}-3z^2)dydz$. $\displaystyle =\int_{-1}^1((2/3)y^{3/2}-3z^2y)\biggr|_{y=0}^{y=1}dz=\int_{-1}^1((2/3)-3z^2)dz$....
Originally Posted by matheagle ???????????? Try quoting the post if its unreadable due to latex errors, at least you will then see what the post contained, and you would not leave the rest of us wondering what you are refering to. CB
I've done that before, but I was too tired and didn't think of doing that tonight.
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