Double integral change of variable polar coordinates question

In summary, the conversation discusses evaluating a double integral using a change of variables and finding the region in polar coordinates. The resulting integral is 1/sqrt(15) * rcosh(2r^2) drdtheta, which is integrated to obtain the answer pi*sinh(2)/4sqrt(15). However, the answer given in the question booklet is pi*sinh(2)/4. This suggests that there may be an error in the question booklet.
  • #1
jamesdocherty
14
0

Homework Statement



evaluate the double integral of cosh(6x^2+10y^2) dxdy by making the change of variables x=rcos(theta)/sqrt(3) and y=rsin(theta)/sqrt(5)

let D be the region enclosed by the ellipse 3x^2+5y^2=1 and the line x=0 where x>0.

Homework Equations

The Attempt at a Solution



first I found the Jacobian, using the determinant of [dx/dr dx/dtheta, dy/dr dy/dtheta] that turned into [cos(theta)/sqrt(3) -rsin(theta)/sqrt(3), sin(theta)/sqrt(5) rcos(theta)/sqrt(5)] to be r/sqrt(15)

then after that i found out what the region in the polar coordinates would be covered, i subbed in x=rcos(theta)/sqrt(3) and y=rsin(theta)/sqrt(5) to find out r is either 1 or -1, as it must be positive it has to be 1. And as it's only half the ellispe on the postive x-axis, theta goes from -pi/2 to pi/2. lastly i subbed in x=rcos(theta)/sqrt(3) and y=rsin(theta)/sqrt(5) into the equation to obtain cosh(2r^2).

Hence now the integral is 1/sqrt(15) * double integral rcosh(2r^2) drdtheta, where 0<r<1 and -pi/2<theta<pi/2 and the 1/sqrt(15) and the r into of cosh is in there due to the jacobian.

integrating this integral i obtained the answer to be pi*sinh(2)/4sqrt(15) which isn't the right answer, i integrated rcosh(2r^2) to be equal to sinh(2r^2)/4 which says is correct on wolfman alpha and then i had 1/sqrt(15)* integral sinh(2)/4 dtheta which turned into 1/sqrt(15)*[theta*sinh(2)/4] for pi/2 and -pi/2 which come out to be pi*sinh(2)/4sqrt(15), i am very confused as why this isn't the right answer and any help would be much appreciated.

MY ANSWER I GOT: pi*sinh(2)/4sqrt(15) (WHICH IS WRONG)
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  • #2
Looks okay to me. Why do you think your answer is wrong?
 
  • #3
Minor point: "intergal" and "intergate" are not words.
 
  • #4
the answer in the question booklet is pi*sinh(2)/4 not pi*sinh(2)/4sqrt(15), do u think there could be an error in the question booklet?
 
  • #5
Yup.
 

Related to Double integral change of variable polar coordinates question

1. What is a double integral in polar coordinates?

A double integral in polar coordinates is a type of integration that is used to find the area of a two-dimensional region in polar coordinates. It is similar to a regular double integral, but the limits of integration and the integrand are expressed in terms of polar coordinates.

2. How do you change the limits of integration for a double integral in polar coordinates?

To change the limits of integration for a double integral in polar coordinates, you need to use the conversion formulas for polar coordinates, which are r = √(x² + y²) and θ = tan⁻¹(y/x). These formulas can be used to express the limits of integration in terms of r and θ.

3. What is the purpose of using polar coordinates in a double integral?

Polar coordinates are often used in double integrals because they can simplify the integrand and the limits of integration. This makes it easier to evaluate the integral and find the area of a region in polar coordinates.

4. How do you set up a double integral problem using polar coordinates?

To set up a double integral problem using polar coordinates, you need to first identify the region in the xy-plane and determine the limits of integration in terms of r and θ. Then, you need to convert the integrand into polar coordinates by substituting x and y with the corresponding polar coordinate formulas. Finally, you can evaluate the integral using standard integration techniques.

5. What are some common applications of double integrals in polar coordinates?

Double integrals in polar coordinates are commonly used in physics, engineering, and other fields to calculate areas, volumes, and other physical quantities. They are also used to find the center of mass of a region and to solve problems involving circular symmetry.

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