Double Integral Problem: Incorrect Jacobian Calculation for Polar Coordinates

[Amaelle]

Homework Statement

integral of int y^2 dxdy over a region (look to the formula inside the exercice)

Relevant Equations

y=rcos(theta)

calculate the double integral

over the region of integration is
x^2 + y^2 ≤ 4; x^2 + (y/4)^2 ≥ 1
the integrals have been made over two regions

my problem is that when I go to the polar coordinate for the ellipsis and use the jacobian i got 2 instead of 8 ( the following is the professor solution)

Any help?
many thanks in advance!

 

[anuttarasammyak]

\int_{A_1}y^2 dx dy = \int_0^{2\pi} d\theta \int_0^2 d\rho\ \rho^3 \sin^2\theta = 16 \int_0^{2\pi} d\theta \int_0^1 d\rho \ \rho^3 \sin^2\theta
A half of it,
\int_{A_2}y^2 dx dy = 8 \int_0^{2\pi} d\theta \int_0^1 d\rho \ \rho^3 \sin^2\theta=\int_{A_1-A_2}y^2 dx dy
What Jacobian you get ?

 

[pasmith]

Homework Statement:: integral of int y^2 dxdy over a region (look to the formula inside the exercice)
Relevant Equations:: y=rcos(theta)

calculate the double integral
<br /> \int y^2\,dx\,dy<br />over the region of integration is
x^2 + y^2 ≤ 4; x^2 + (y/4)^2 ≥ 1
the integrals have been made over two regions

my problem is that when I go to the polar coordinate for the ellipsis and use the jacobian i got 2 instead of 8 ( the following is the professor solution)

Any help?
many thanks in advance!

For the ellipse, you need to use the parametrization <br /> \begin{align*} x &amp;= \rho \cos \vartheta, \\ y &amp;= 2\rho\sin \vartheta. \end{align*} With this choice, the ellipse is at \rho = 1, rather than the complicated function of \vartheta you would get from actual polar coordinates.

You should obtain a factor of 2 from the jacobian and a further two factors from the integrand.

 

[Amaelle]

Thanks a lot
so first change
u=x
v=y/2
so the jacobian equal 2
after
v=2rsin(theta)
v^2=4[sin(theta)]^2

I got it now

 

[Amaelle]

\int_{A_1}y^2 dx dy = \int_0^{2\pi} d\theta \int_0^2 d\rho\ \rho^3 \sin^2\theta = 16 \int_0^{2\pi} d\theta \int_0^1 d\rho \ \rho^3 \sin^2\theta
A half of it,
\int_{A_2}y^2 dx dy = 8 \int_0^{2\pi} d\theta \int_0^1 d\rho \ \rho^3 \sin^2\theta=\int_{A_1-A_2}y^2 dx dy
What Jacobian you get ?

Thanks a lot!