∫∫ x^2 dA ; bounded by ellipse

In summary, the problem is to calculate the integral ∫∫ x^2 dA, bounded by the ellipse 9x^2+4y^2=36. Two methods were attempted: polar coordinate conversion and u,v variable change. The first method yielded the correct answer of 6π after computing the Jacobian. The second method, however, had incorrect ranges and did not take into account the elliptical shape, resulting in an incorrect answer of 32. The correct answer is 6π.
  • #1
red6290
4
0

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.
 
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  • #2
Try 3x = r cos(theta), and 2y = r cos(theta), with r=[0,6] and theta=[0,2pi]... It might work.
 
  • #3
∫∫ 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?
 
  • #4
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?
 
  • #5


red6290 said:
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 [itex]\sqrt{1-u^2}[/itex]. i.e.,

[tex]2\int_{-1}^1 du u \int_{0}^{\sqrt{1-u^2}}dv[/tex]

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 [itex]dxdy = r dr d\theta[/itex]. 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 [itex]6\pi[/itex].
 
Last edited:
  • #6


Mute said:
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 [itex]dxdy = r dr d\theta[/itex]. 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 [itex]6\pi[/itex].

thank you very much it was really helpful!
 
  • #7
(two threads merged)
 

1. What is the meaning of the notation "∫∫ x^2 dA"?

The notation "∫∫ x^2 dA" represents a double integral, where x^2 is the function being integrated and dA represents the area element. This integral is also known as a surface integral, as it calculates the volume under a surface.

2. What is the significance of the ellipse in this integral?

The ellipse serves as the boundary for the region of integration. This means that the integral is only calculated for values of x and y that fall within the ellipse's boundaries.

3. How is the region of integration determined?

The region of integration is determined by the intersection of the ellipse and the xy-plane. This intersection creates a closed shape that is used to define the limits of the integral.

4. Can the ellipse be replaced with another shape for the region of integration?

Yes, the ellipse can be replaced with any shape as long as it can be described by a set of equations. This could include circles, squares, or even more complex shapes.

5. What is the purpose of calculating this double integral?

The purpose of calculating this double integral is to find the volume under the surface defined by the function x^2 within the boundaries of the ellipse. This is a useful tool in many scientific and mathematical applications, such as calculating the volume of a three-dimensional object or finding the mass of a curved surface.

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