Evaluating a Double Integral Using Polar Coordinates

[pern_comoto]

Okay I have no idea where to start on this example problem:

Use polar coordinates to evaulate the double integral e^((x^2)+(y^2))dydx
[frist (inner) integal lower limit y= -sqrt(4-x^2) upper limit y=0)]
[second (outer) lower limit x=0 upper limit x=2]

When I start doing the integral of e^((x^2)+(y^2))dy I get some really crazy answer and then I don't know if I should put it in polar coordinates before I try and take the integral or after. Can you tell me where to start?

 

[HallsofIvy]

Okay I have no idea where to start on this example problem:

Use polar coordinates to evaulate the double integral e^((x^2)+(y^2))dydx
[frist (inner) integal lower limit y= -sqrt(4-x^2) upper limit y=0)]
[second (outer) lower limit x=0 upper limit x=2]

When I start doing the integral of e^((x^2)+(y^2))dy I get some really crazy answer and then I don't know if I should put it in polar coordinates before I try and take the integral or after. Can you tell me where to start?

How in the world would you get "some really crazy answer"? It's pretty well known that there is no elementary anti-derivative for $e^{-x^2}$- not even a "crazy" one!

Also, since the problem is to evaluate the integral, I would see no point in changing to polar coordinates after integrating!

What do you start? By doing what the problem says: "use polar coordinates"!
Of course $e^{-(x^2+ y^2)}$ converts to $e^{-r^2}$. Do you know how dydx converts?

To get the limits of integration, draw a picture. $y= -\sqrt{4- x^2}$ is the lower half of the circle $x^2+ y^2= 4$ which has center at the origin and radius 2. x going from 0 to 2 means you are to the right of the y-axis. The region you are integrating over is the lower right quarter of a circle of radius 2. How do r and $\theta$ change to cover that region?