Use polar coordinates to evaluate the double integral.

Question

anonymous · Answer

$$\int\limits_{?}^{?} \int\limits_{?}^{?}f(x,y) dA$$ where  $$f(x,y)=e ^{-(x^2+y^2)} $$ and $$ R=\left\{ (x,y)/x^2+y^2\le4,x \ge0,y \ge0 \right\}$$

anonymous · Answer

The double integrals are over the region R

turingtest · Answer

I thought I did this problem yesterday...
was that with someone else, or did you just not like my answer?

anonymous · Answer

Probably someone else... If you had a link that would be helpful and then I could just ask any question I had here or there.

anonymous · Answer

Maybe they are in my class.

turingtest · Answer

nah it's okay...
do you know the formulas for conversion to polar coordinates?

turingtest · Answer

...or how about the shape of the region R ?
those things make this problem easy

turingtest · Answer

what is$$x^2+y^2=4$$?

anonymous · Answer

a circle

turingtest · Answer

|dw:1334058440796:dw|it is a circle of radius 2, right?
so in polar coordinates the bounds on this are pretty obvious
but there is one addition thing here :$$x,y\ge0$$so what are our bounds for the radius and angle in this case?

turingtest · Answer

talk to me
if you are confused about something, please say so...

anonymous · Answer

attachment

turingtest · Answer

@eliassaab could you please not post the answer?

anonymous · Answer

I am not

turingtest · Answer

ok, couldn't open the file so I didn't know
thanks

anonymous · Answer

well it is a quarter circle, so we are integrating from 0 to 2

turingtest · Answer

awesome, and the angle?

anonymous · Answer

When we put it in polar coordinates won't that just be a term in the integral until we get to the end when we evaluate it from 0 to 2pi right?

turingtest · Answer

that would be the whole circle, but $$x,y\ge0$$implies that we are in quadrant 1(you yourself said it was a quarter of a circle)
so the angle only goes to...?

anonymous · Answer

pi/2

turingtest · Answer

yes
so now do the polar substitutions for x, y, and dA
do you know them?$$x=?,y=?,dA=?$$

anonymous · Answer

x=r cos 	heta  
y=r sin 	heta  
dxdy=rdrd 	heta

turingtest · Answer

perfect!
now just plug in, simplify, and integrate :)

anonymous · Answer

Does this look right to you?
$$\int\limits_{0}^{\pi/2}\int\limits_{0}^{2}re^{-r^2}drd \Theta=(e^4-1)\pi/(4e^4)$$

turingtest · Answer

yeah looks good at first glance
let me make sure though

turingtest · Answer

oh yeah after simplifying, lol
that confused me a little at first
I would write it$$\frac\pi4(1-e^{-4})$$but yeah, that's right I believe.

anonymous · Answer

Ah yeah I guess that might be a little cleaner notation.

turingtest · Answer

anyway, good job, hope that helped :)