   Chapter 15.9, Problem 17E

Chapter
Section
Textbook Problem

Use the given transformation to evaluate the integral.17. ∫∫R x2 dA, where R is the region bounded by the ellipse 9x2 + 4y2 = 36; x = 2u, y = 3v

To determine

To evaluate: The integral Rx2dA.

Explanation

Given:

The region R is bounded by the ellipse 9x2+4y2=36 and x=2u, y=3v

Property used: Change of Variable

Change of Variable in double integral is given by,

Rf(x,y)dA=Sf(x(u,v),y(u,v))|(x,y)(u,v)|dudv (1)

Formula used:

If g(x) is the function of x and h(y) is the function of y then,

ababg(x)g(y)dydx=abg(x)dxabh(y)dy (2)

Calculation:

Obtain the Jacobian, (x,y)(u,v)=|xuxvyuyv|

Find the partial derivative of x and y with respect to u and v. x=2u then xu=2 and xv=0 and y=3v then yu=0 and yv=3.

(x,y)(u,v)=|2003|=2(3)(0)(0)=60=6

From the given integral the function is, x2 and substitute the values of x and y.

x2=(2u)2=4u2

Find the boundary by using the given transformation.

The ellipse (3x)2+(2y)2(6)2 is the image of u2+v21

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