Double Integral Problem: Incorrect Jacobian Calculation for Polar Coordinates

In summary, the problem involves calculating a double integral over a region of integration that consists of two regions. The difficulty lies in converting the integral to polar coordinates and using the appropriate Jacobian. After some calculations, a factor of 2 is obtained from the Jacobian and a further two factors from the integrand, leading to the correct solution.
  • #1
Amaelle
310
54
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
1597319551559.png


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
1597319752949.png


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)
1597319891086.png


Any help?
many thanks in advance!
 
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  • #2
[tex]\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[/tex]
A half of it,
[tex]\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[/tex]
What Jacobian you get ?
 
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  • #3
Amaelle said:
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
[tex]
\int y^2\,dx\,dy
[/tex]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 [tex]
\begin{align*} x &= \rho \cos \vartheta, \\ y &= 2\rho\sin \vartheta. \end{align*}[/tex] With this choice, the ellipse is at [itex]\rho = 1[/itex], rather than the complicated function of [itex]\vartheta[/itex] 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.
 
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Likes Amaelle
  • #4
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
 
  • #5
anuttarasammyak said:
[tex]\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[/tex]
A half of it,
[tex]\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[/tex]
What Jacobian you get ?
Thanks a lot!
 

Related to Double Integral Problem: Incorrect Jacobian Calculation for Polar Coordinates

1. What is a double integral?

A double integral is a type of mathematical operation that involves calculating the area under a two-dimensional function. It is represented by the symbol ∫∫ and is used to find the volume between a surface and a given region in the x-y plane.

2. What are the applications of double integrals?

Double integrals have various applications in physics, engineering, and economics. They are used to calculate the mass, center of mass, and moments of inertia of a two-dimensional object. They are also used in calculating the work done by a variable force and in finding the probability density of a continuous random variable.

3. What is the difference between a single and a double integral?

A single integral is used to find the area under a one-dimensional function, while a double integral is used to find the volume under a two-dimensional function. In a single integral, the limits of integration are on a single axis, while in a double integral, the limits of integration are on two axes.

4. How do I solve a double integral?

To solve a double integral, you need to first identify the limits of integration and the order of integration. Then, you can use various integration techniques, such as substitution, integration by parts, or partial fractions, to evaluate the integral. It is also important to sketch the region of integration to visualize the problem and make the integration process easier.

5. What are some common challenges in solving double integrals?

Some common challenges in solving double integrals include determining the correct limits of integration, choosing the appropriate order of integration, and handling complex integrands. It is also important to be familiar with various integration techniques and have a good understanding of the properties of double integrals.

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