MHB Change of Variables in Multiple Integrals

[harpazo]

Let S S = double integrals

S S x^2 dA; where R is the region bounded by the ellipse
9x^2 + 4y^2 = 36.

The given transformation is x = 2u, y = 3v

I decided to change the given ellipse to a circle centered at the origin.

9x^2 + 4y^2 = 36

I divided across by 36.

x^2/4 + y^2/9 = 1

I replaced x with 2u and y with 3v.

I got u^2 + v^2 = 1.

I let u^2 + v^2 = r^2.

r^2 = 1

sqrt{r^2} = sqrt{1}

r = 1

This tells me the bounds of r to be 0 to 1. The bounds of theta are 0 to 2pi.

dA = rdrdθ

The function x^2, I replaced by (2u)^2 = 4u^2.

I found the Jacobian to be 6.

The double integral set I came up with is

S S 4u^2 6 rdrdθ, where the inner bounds are r = 0 to r = 1 and the outer bounds are θ = 0 to θ = 2pi.

I then pulled out the constant 6.

6 S S 4u^2 r drdθ

When I evaluate the double integral, I cannot finish the problem. Shouldn't rdr be replaced with du? If this true, then the inner bounds should be in terms of u not r. I would like someone to tell me where I went wrong and to set up the correct double integrals for this problem. Do not evaluate the integrals.

 

[Joppy]

Let S S = double integrals

$\int \int x^2 dA$; where $R$ is the region bounded by the ellipse
$9x^2 + 4y^2 = 36$.

The given transformation is $x = 2u$, $y = 3v$

I decided to change the given ellipse to a circle centered at the origin.

$9x^2 + 4y^2 = 36$

I divided across by $36$.

$x^{\frac{2}{4}} + y^{\frac{2}{9}} = 1$

I replaced $x$ with $2u$ and $y$ with $3v$.

I got $u^2 + v^2 = 1$.

I let $u^2 + v^2 = r^2$.

$r^2 = 1$

$\sqrt{r^2} = \sqrt{1}$

$r = 1$

This tells me the bounds of $r$ to be $0$ to $1$. The bounds of theta are $0$ to $2 \pi$.

$dA = rdrdθ$

The function $x^2$, I replaced by $(2u)^2$ = $4u^2$.

I found the Jacobian to be 6.

The double integral set I came up with is

$\int \int 4u^2 6 r \ drdθ$, where the inner bounds are $r = 0 \ to \ r = 1$ and the outer bounds are $θ = 0 \ to \ θ = 2pi$.

I then pulled out the constant 6.

$6 \int \int 4u^2 r dr dθ$

When I evaluate the double integral, I cannot finish the problem. Shouldn't rdr be replaced with du? If this true, then the inner bounds should be in terms of u not r. I would like someone to tell me where I went wrong and to set up the correct double integrals for this problem. Do not evaluate the integrals.

Just improving the readability a bit.

 

[harpazo]

I found the answer. I let u = rcos(theta). I got 6pi.