Question

Use polar coordinates to set up and evaluate the double integral:

e^-(x^2+y^2)/2

Answer 1

\[f(x,y)=e^-(x^2+y^2), R: x^2+y^2\le25,x \ge0\]

Answer 2

For the limits I got \[y=\sqrt{x^2-5}\], y = 0, x=5, x=0

Answer 3

So, \[\int\limits_{0}^{5}\int\limits_{0}^{\sqrt{x^2-5}}e^{-(x^2+y^2)/2}dydx\]

Answer 4

This is how far I've gotten right now, I think next I have to convert the integrals to polar coordinates.

Answer 5

let x^+y^ = r^2
and
dxdy = r dr d\(\theta\)
It should be pretty easy now.

Answer 6

alright, I'll try solving this and let you guys know what I get or if I get stuck somewhere.

Answer 7

^ Since I'm converting the limits would change right? If so would they be \[d \theta=0,pi/2, r=0,5\]

Answer 8

dθ=-pi/2,pi/2,r=0,5

Answer 9

Oh okay...

Answer 10

\[\int\limits_{-pi/2}^{pi/2}\int\limits_{0}^{5}r^2(\cos^2\theta+\sin^2\theta)r dr d \theta\]
\[1/8\int\limits_{-pi/2}^{pi/2}-r ^{4} d \theta\]

\[1/8\int\limits_{-pi/2}^{pi/2}625d \theta\]
^I got this by plugging in 5 and 0.

Answer 11

That's what I get when I evaluated with respect to r. I have a feeling that I should of brought that - sign over with the 1/8

Answer 12

\[e^{(x^2+y^2)/2}\]
or
\[e^{x^2+y^2}/2\] ?
@JerJason

Answer 13

The first one, sorry I forgot to change that when I originally posted it

Answer 14

ah ok...

so you see:
0<=x<=5
-sqrt(25-x^2)<=y<=sqrt(25-x^2)
so we'll have the double integral
\[\int\limits_{0}^{5} \int\limits_{-\sqrt{25-x^2}}^{\sqrt{25-x^2}}e^{-(x^2+y^2)/2}dydx\]
now looking at this with polar coordinates....

we can see that
-pi/2<=theta<=pi/2
0<=r<=5
recall that x^2+y^2=r^2 so
e^(-(x^2+y^2)/2)=e^(-r^2/2)
so we'll have the double integral:
\[\int\limits_{-pi/2}^{pi/2} \int\limits_{0}^{5}e^{-r^2/2}rdrd \theta\]