∫∫ x^2 dA ; bounded by ellipse

[red6290]

Homework Statement

try ∫∫G x^2 dA ;value is the region bounded by the ellipse 9x^2+4y^2=36

Homework Equations

The Attempt at a Solution

i think i have to change the variables to polar coordinate or U,V function but i have no idea.

 

[Char. Limit]

Try 3x = r cos(theta), and 2y = r cos(theta), with r=[0,6] and theta=[0,2pi]... It might work.

 

[red6290]

∫∫ x^2 dA ; bounded by ellipse; which soln is right?

Homework Statement

the problem is:

try ∫∫ x^2 dA ; bounded by ellipse 9x^2+4y^2=36


and i tried 2 different ways but i don't know which one is right.

Homework Equations

The Attempt at a Solution

first method: polar coordinate conversion.

3x= r cos (theta), 2y = r sin (theata) 0 < r < 6 , 0 < theta < 2 pi

so ∫∫ r^2 (cos(theta))^2 r dr d(theta)

= 36 pi



second method: u,v variable change

9x^2+4y^2 = 36 -> x^2/4 + y^2/9 = 1

u = x/2 , v = y/3
(-1 < u < 1, -1 < v < 1) ( I am not sure about this range. is this right?, if not what is wrong with this?)

so x= 2u

∫∫ x^2 dA = ∫∫ 4 u^2 (Jacobian=6) ds = ∫∫ 24 u^2 du dv= 32


36 pi vs. 32 ; which is right?

 

[red6290]

thank you for the answer but i have another problem.

i tried 2 different ways but i don't know which one is right.


first method: polar coordinate conversion.

3x= r cos (theta), 2y = r sin (theata) 0 < r < 6 , 0 < theta < 2 pi

so ∫∫ r^2 (cos(theta))^2 r dr d(theta)

= 36 pi



second method: u,v variable change

9x^2+4y^2 = 36 -> x^2/4 + y^2/9 = 1

u = x/2 , v = y/3
(-1 < u < 1, -1 < v < 1) ( I am not sure about this range. is this right?, if not what is wrong with this?)

so x= 2u

∫∫ x^2 dA = ∫∫ 4 u^2 (Jacobian=6) ds = ∫∫ 24 u^2 du dv= 32


36 pi vs. 32 ; which is right?

 

[Mute]

second method: u,v variable change

9x^2+4y^2 = 36 -> x^2/4 + y^2/9 = 1

u = x/2 , v = y/3
(-1 < u < 1, -1 < v < 1) ( I am not sure about this range. is this right?, if not what is wrong with this?)

so x= 2u

∫∫ x^2 dA = ∫∫ 4 u^2 (Jacobian=6) ds = ∫∫ 24 u^2 du dv= 3236 pi vs. 32 ; which is right?

Your range is not correct. The ellipse in the x-y plane is a circle in the u-v plane. The range -1 to 1 for u and v describes a square in the u-v plane. To really describe the circle one of the limits needs to depend on the other variable. For example, noting that the integrand u^2 is invariant if we shift the angle by pi, we can just do the integral over the top half of the circle and double the result. So, if u runs from -1 to 1, you need an integral in v that runs from zero up to $\sqrt{1-u^2}$. i.e.,

$$2\int_{-1}^1 du u \int_{0}^{\sqrt{1-u^2}}dv$$

The resulting integral in u will require you to make a trigonometric substitution.

The Attempt at a Solution

first method: polar coordinate conversion.

3x= r cos (theta), 2y = r sin (theata) 0 < r < 6 , 0 < theta < 2 pi

so ∫∫ r^2 (cos(theta))^2 r dr d(theta)

= 36 pi

This method is almost correct. You forgot to compute the Jacobian. It's not simply $dxdy = r dr d\theta$. There's an extra factor of 1/6 that comes out in the Jacobian because you have an ellipse instead of just a circle. The proper answer is $6\pi$.

 

[red6290]

Your range is not correct. The ellipse in the x-y plane is a circle in the u-v plane. The range -i to 1 for u and v describes a square in the u-v plane. So, this method is not correct.



This method is almost correct. You forgot to compute the Jacobian. It's not simply $dxdy = r dr d\theta$. There's an extra factor of 1/6 that comes out in the Jacobian because you have an ellipse instead of just a circle. The proper answer is $6\pi$.

thank you very much it was really helpful!

 

[berkeman]

(two threads merged)