Triple Integral Cartesian Coordinates

[bytenel]

Ok I have a quick question. I have this problem that is doable with polar coordinates and triple integrals but I was wondering if it would be possible to do this problem in the cartesian coordinate system (odd question I know...).

Homework Statement

A sprinkler distributes water in a circular pattern, supplying water to a depth of e^(-r) feet per hour at a distance of r feet from the sprinkler.

A. What is the total amount of water supplied per hour inside of a circle of radius 10?
2pi-2pie^(-10)
B. What is the total amount of water that goes through the sprinkler per hour?
2pi

Homework Equations

Just integration techniques I guess. pi*R^2 is the equation for a circle area. x^2+y^2=100 is the equation in standard form for this circle.

The Attempt at a Solution

Here's where I get lost. In cartesian coordinates the bounds for the resulting double integral should be 0<=y<=sqrt(100-x^2) and 0<=x<=10, right? Then from there I take the double integral of the equation of the circle and...?

Any help is appreciated, thanks!

EDIT: Should I take the double integral of pi(x^2+y^2) with the bounds I have above? or is that wrong?

 

[LCKurtz]

Here's where I get lost. In cartesian coordinates the bounds for the resulting double integral should be 0<=y<=sqrt(100-x^2) and 0<=x<=10, right? Then from there I take the double integral of the equation of the circle and...?

Any help is appreciated, thanks!

EDIT: Should I take the double integral of pi(x^2+y^2) with the bounds I have above? or is that wrong?

Those bounds would be for a quarter of the circle. You could use those limits and multiply the answer by 4. But your integrand would be $e^{-r}=e^{-\sqrt{x^2+y^2}}$. I don't think anyone would want to work it in rectangular coordinates. If you really must try it, you might have success with a trig substitution on the inside integral. Not sure, I haven't tried it.

 

[bytenel]

Wait, why would those limits only be for a quarter of the circle?

 

[LCKurtz]

Wait, why would those limits only be for a quarter of the circle?

Because your limits have both x and y positive.