# 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

Staff Emeritus
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!