Converting to Polar Coordinates

[1question]

Homework Statement

Convert ∫ from 0 to 3/√2 ∫ from y to √(9-y^2) of xydxdy to polar form.

Homework Equations

x²+y²=r²

The Attempt at a Solution

I found the equation x²+y^2=9 from the upper range of the second integral. So r=3. Therefore r ranges from 0 to 3. The integrand is now (rcosθ)(rsinθ)rdrdθ.

I'm having trouble with the range of integration of the first integral (0 to 3/√2). From the other information, I think that the area I'm integrating is from -π/2 to π/2. This is wrong. Any hints? Thanks.

 

[Mark44]

Homework Statement

Convert ∫ from 0 to 3/√2 ∫ from y to √(9-y^2) of xydxdy to polar form.

Homework Equations

x²+y²=r²

The Attempt at a Solution

I found the equation x²+y^2=9 from the upper range of the second integral. So r=3. Therefore r ranges from 0 to 3. The integrand is now (rcosθ)(rsinθ)rdrdθ.

I'm having trouble with the range of integration of the first integral (0 to 3/√2). From the other information, I think that the area I'm integrating is from -π/2 to π/2. This is wrong. Any hints? Thanks.

It's essential in these kinds of problems to have a good understanding of what the region of integration looks like. Can you describe, in words, what this region looks like?
What does the lower limit of integration, y, represent?

 

[1question]

Oh. Is it the straight line x=y?

The region I've got so far is a semi-circle from -pi/2 to pi/2. Not really sure how to deal with the x=y.

 

[Mark44]

Here's your integral:
$$ \int_{y = 0}^{3/\sqrt{2}} \int_{x = y}^{\sqrt{9 - y^2}}~xy~dx~dy$$
If you click on the integral, you can see the LaTeX I wrote to produce it.
The inner integration goes across, from left to right, from the line to your half circle. The outer integration goes up from y = 0 to y = 3/√2.

You should draw a sketch of this region.