- #1
Gauss M.D.
- 153
- 1
Homework Statement
Calculate
[itex]\int \int x dx dy[/itex]
Over the area defined by [itex]1 \leq x^{2} + 4y^{2} \leq 9[/itex]
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 [itex]1 \leq u^{2} + v^{2} \leq 9, u,v\geq0 [/itex]
and the integral
[itex]\sqrt{3} \int \int u - \frac{1}{\sqrt{3}} v du dv[/itex]
Now we will switch to polar:
[itex]u^{2} + v^{2} = ρ^{2}[/itex]
[itex]u = ρ cos(θ)[/itex]
[itex]v = ρ sin(θ)[/itex]
Since the functional determinant is ρ, this gives us
[itex]\sqrt{3} \int \int ρ(ρ cos(θ) - \frac{1}{\sqrt{3}} ρ sin(θ))dρ dθ[/itex]
Since [itex] 1 \leq u^{2} + v^{2} \leq 9[/itex] and [itex]x,y \geq 0[/itex]:
[itex]1 \leq ρ \leq 3, 0 \leq θ \leq \frac{\pi}{2}[/itex]
[itex]\sqrt{3} /3 \int [ρ^{3} cos(θ) - \frac{1}{\sqrt{3}} ρ^{3} sin(θ)] dθ[/itex]
[itex]\sqrt{3} /3 \int [27 cos(θ) - \frac{27}{\sqrt{3}} sin(θ) - cos(θ) + \frac{1}{\sqrt{3}} sin(θ)] dθ[/itex]
[itex]\sqrt{3} /3 \int 26 cos(θ) - \frac{26}{\sqrt{3}} sin(θ) dθ[/itex]
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.