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Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347

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BuyFindarrow_forward

Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347
Textbook Problem

Finding Volume Using a Change of Variables In Exercises 23-30, use a change of variables to find the volume of the solid region lying below the surface z = f ( x , y ) and above the plane region R.

f ( x , y ) = ( 3 x + 2 y ) 2 2 y x

R: region bounded by the parallelogram with vertices (0, 0), (-2, 3), (2, 5), (4, 2)

To determine

To calculate: The volume of the provided solid region lying below the surface z=(3x+2y)22yx and above the plane region R.

Explanation

Given:

The provided function is f(x,y)=(3x+2y)22yx.

The region R by parallelogram with vertices (2,5),(4,2),(0,0) and (2,3).

Formula used:

Jacobians formula, (x,y)(u,v)=|xvxuyuyv|

And change of variables for double integrals is given as

Rf(x,y)dxdy=Sf(g(u,v),h(u,v))|(x,y)(u,v)|dudv

The slope-intercept form of equation of line y=mx+c, where ‘m’ is the slope of the line and m=y2y1x2x1.

Calculation:

The region R by parallelogram with vertices (2,5),(4,2),(0,0) and (2,3). By using given coordinates, plot the parallelogram.

The slope of the straight line between the point (0,0) and (2,3) is calculated as,

m=3020=32

The straight-line equation between the point (0,0) and (2,3) is calculated as,

y0=32(x0)2y=3x3x+2y=0

The slope of straight line between the point (0,0) and (4,2) is calculated as,

m=2040=12

The straight-line equation between the point (0,0) and (4,2) is calculated as,

y0=12(x0)2y=x2yx=0

The slope of straight line between the point (2,5) and (4,2) is calculated as,

m=2542=32

The straight-line equation between the point (2,5) and (4,2) is calculated as,

y2=32(x4)2y4=3x+123x+2y=16

The slope of straight line between the point (2,5) and (2,3) is calculated as,

m=3522=24=12

The straight-line equation between the point (2,5) and (2,3) is calculated as,

y3=12(x+2)2y6=x+22yx=8

Assume that 2yx=u and 3x+2y=v.

Subtracting these two equations, we get,

3x+x=vux=vu4

Put x=vu4 in the second equation,

2y(vu4)=u2y=v+3u4y=v+3u8

The line of the equation 2yx=0 in new coordinate (u,v) is given as,

2yx=0u=0

The line of the equation 3x+2y=0 in new coordinate (u,v) is given as,

3x+2y=0v=0

The line of the equation x2y=8 in new coordinate (u,v) is given as,

2yx=8u=8

The line of the equation 3x+2y=0 in new coordinate (u,v) is given as,

3x+2y=16v=16

The straight line in a new rectangular co-ordinate is shown as below

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Chapter 14 Solutions

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