Homework Help: Double integral substitution

1. Mar 4, 2014

Swasse

Hey.

1. The problem statement, all variables and given/known data

∫∫x^3 dxdy, with the area of integration: D={(x,y)∈R^2: 1<=x^2+9y^2<=9, x>=3y}

3. The attempt at a solution

Did the variable substitution u=x and v=3y so the area of integration became 1<=u^2 + v^2 <=9, u>=v. And the integral became ∫∫(1/3)u^3 dudv. Then I switched to polar coordinates and the integral ∫∫(1/3)r^4cos^3θ drdθ, with 1<=r<=9 and -3/4<=θ<=pi/4.

Any errors so far?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Mar 4, 2014
2. Mar 4, 2014

HallsofIvy

Well, first, if you take u= x, v= 3y then 1<= x^2+ y^2<= 9 does NOT become 1<= u^2+ v^2<= 9, it becomes 1<= u^2+ v^2/9<= 9, the region between two ellipses. Since you then change to polar coordinates, I don't see any good reason to do that. Rather, change to polar coordinates initially. The line x= 3y, or y= x/3 has $\theta= arctan(1/3)$ so the integral is $$\int_{\theta= -\pi+ arctan(1/3)}^{arctan(1/3)}\int_{r= 1}^3 r^4 cos^3(\theta) drd\theta$$.

3. Mar 4, 2014

Swasse

My mistake, it should have said 1<= x^2+ 9y^2<= 9

4. Mar 5, 2014

Swasse

Can someone please confirm if my way of solving is correct? I made a typo in the first post, but I have corrected it now.

5. Mar 5, 2014

vela

Staff Emeritus
Your upper limit for $r$ is wrong.

6. Mar 5, 2014

Swasse

I had a feeling it was, because my answer was in the right form, just too large. Can you please explain why it is wrong?

7. Mar 5, 2014

Swasse

Just figured it out, the radius is not 9, it is √9=3. Is that the correct limit?

8. Mar 5, 2014

vela

Staff Emeritus
Yup!