Question

f(x,y) = (5/2)e^(x^2+y^2)

dydx

from y = -sqrt(4-x^2) to y = sqrt(4-x^2)
from x = - 2 to x = 2

Answer 1

I need help integrating.

Answer 2

I know the answer, confirmed by Wolfram Alpha and the solutions guide, I just don't know how to get there.

Answer 3

Have you converted to polar?

Answer 4

Cylindrical coordinates make it easier but I would like to keep it in cartesian.

Answer 5

then:
\(5/2\int_{-\sqrt(4-x^2)}^{\sqrt(4-x^2)}\int_{-2}^{2}e^{x^2}e^{y^2}dxdy\)

Answer 6

without an extra y or x, how do you differentiate e^(x^2) or e^(y^2)?

Answer 7

A change in coordinates will simplify the work.
\[x=r\cos\theta\\
y=r\sin\theta\\
x^2+y^2=r^2\\
dy~dx=r~dr~d\theta\]
Then change the limits accordingly.

Answer 8

Thanks Siths but the question wants to see both versions. It's goal is to demonstrate how changing coordinate systems simplifies things.

Answer 9

Well @myko's integral gives you the Cartesian integral. Apply the conversions and you'll get your polar integral.

Answer 10

They can't really expect you to evaluate the Cartesian one, can they?

Answer 11

you will not find it easy to integrate in cartesian, Look about erf(x) function in wikipedia

Answer 12

I'll ask the instructor because we never learned the error function.

Answer 13

\[\frac{5}{2}\int_{-\sqrt{4-x^2}}^\sqrt{4-x^2}\int_{-2}^2e^{x^2+y^2}~dy~dx=\frac{5}{2}\color{red}{\int_{0}^{2\pi}}\color{blue}{\int_{0}^{2}}e^{r^2}~r~\color{blue}{dr}~\color{red}{d\theta}\]

Answer 14

Should be \(dx~dy\), but you get the idea...

Answer 15

I think you are missing an r to account for the Jacobian transformation but I got it. Thanks.

Answer 16

No, it's there. \(dx~dy=\color{red}{r}~dr~d\theta\).