# Multiple integrals

• May 25th 2009, 09:49 PM
Multiple integrals
Question # 01:

Evaluate the iterated integral by converting to polar coordinates

http://www.mathhelpforum.com/math-he...1&d=1243328603

The region of integration is bounded by
http://www.mathhelpforum.com/math-he...1&d=1243316586

Question # 02:

Evaluate the triple integral http://www.mathhelpforum.com/math-he...1&d=1243316757

http://www.mathhelpforum.com/math-he...1&d=1243316882

• May 25th 2009, 09:58 PM
matheagle
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 $\displaystyle u=r^2$.
• May 25th 2009, 10:01 PM
Quote:

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.

$\displaystyle \int_{-0}^{2}\int_{-(4-x^2)^{1/2}}^{(4-x^2)^{1/2}} e^{-x^2-y^2}dy dx$
• May 25th 2009, 10:08 PM
matheagle
The second one is

$\displaystyle \int_2^3dx\int_{-1}^1\int_0^1 (\sqrt{y}-3z^2)dydz=\int_{-1}^1\int_0^1(\sqrt{y}-3z^2)dydz$.

$\displaystyle =\int_{-1}^1((2/3)y^{3/2}-3z^2y)\biggr|_{y=0}^{y=1}dz=\int_{-1}^1((2/3)-3z^2)dz$....
• May 25th 2009, 11:13 PM
CaptainBlack
Quote:

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
• May 25th 2009, 11:53 PM
matheagle
I've done that before, but I was too tired and didn't think of doing that tonight.