Double Integral over Elliptical Area: Polar Coordinates and Substitution Method

[Gauss M.D.]

Homework Statement

Calculate

$\int \int x dx dy$

Over the area defined by $1 \leq x^{2} + 4y^{2} \leq 9$

Homework Equations

The Attempt at a Solution

First we'll do the sub:

u = x + y
v = sqrt(3)y

Which gives us the area $1 \leq u^{2} + v^{2} \leq 9, u,v\geq0$

and the integral

$\sqrt{3} \int \int u - \frac{1}{\sqrt{3}} v du dv$

Now we will switch to polar:

$u^{2} + v^{2} = ρ^{2}$
$u = ρ cos(θ)$
$v = ρ sin(θ)$

Since the functional determinant is ρ, this gives us

$\sqrt{3} \int \int ρ(ρ cos(θ) - \frac{1}{\sqrt{3}} ρ sin(θ))dρ dθ$

Since $1 \leq u^{2} + v^{2} \leq 9$ and $x,y \geq 0$:
$1 \leq ρ \leq 3, 0 \leq θ \leq \frac{\pi}{2}$

$\sqrt{3} /3 \int [ρ^{3} cos(θ) - \frac{1}{\sqrt{3}} ρ^{3} sin(θ)] dθ$

$\sqrt{3} /3 \int [27 cos(θ) - \frac{27}{\sqrt{3}} sin(θ) - cos(θ) + \frac{1}{\sqrt{3}} sin(θ)] dθ$

$\sqrt{3} /3 \int 26 cos(θ) - \frac{26}{\sqrt{3}} sin(θ) dθ$

Evaluating this between 0 and pi/2 nets something like 26 sqrt(3) blabla. Something terribly off. What am I doing wrong? Been wrestling for hours with the same problem now.

 

[davidchen9568]

Why not u=x, v=2y?

 

[Gauss M.D.]

Goddamnit.

Because the region is bounded by x^2 + 4y^2 + 2xy. Typo...

 

[haruspex]

Since $1 \leq u^{2} + v^{2} \leq 9$ and $x,y \geq 0$:
$1 \leq ρ \leq 3, 0 \leq θ \leq \frac{\pi}{2}$

I don't see any previous mention of $x,y \geq 0$. Are you sure that's part of the boundary? And the map to a range for θ is not that simple - remember ρ and θ relate to x and y via u and v.

 

[Gauss M.D.]

Yeah, x and y are both positive, which means u and v are too.

I am guessing we already identified the issue though. I think I dropped the 2xy somewhere when trying to express the integrand x in u and v... Gonna check.

 

[Ray Vickson]

Goddamnit.

Because the region is bounded by x^2 + 4y^2 + 2xy. Typo...

Could you please re-write the whole question exactly? I can't figure out what you mean here.

 

[haruspex]

Yeah, x and y are both positive, which means u and v are too.

I am guessing we already identified the issue though. I think I dropped the 2xy somewhere when trying to express the integrand x in u and v... Gonna check.

You didn't answer my second point, that the region 1 ≤ ρ ≤ 3, 0 ≤ θ ≤ π/2 is the same as u ≥ 0, v ≥ 0, but it's not the same as x ≥ 0, y ≥ 0. E.g. consider x = -1, y = 1.