- #1

jamesdocherty

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## 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.

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:

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MY ANSWER I GOT:

**pi*sinh(2)/4sqrt(15) (WHICH IS WRONG)**
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