   Chapter 7.9, Problem 12E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Comparing Different Orders of Integration In Exercises 7-12, set up the integrals for both orders of integration and use the more convenient order to find the volume of the solid region bounded by the surface f(x, y) and the planes. See Example 2. f ( x , y ) = 1 2 x 2 + 1 Planes: z = 0 , y = 0 , y = 3 x , x = 4

To determine

To calculate: The volume of solid region bounded by surface f(x,y)=12x2+1 and plane: z=0,y=0,y=3x,x=4.

Explanation

Given information:

The provided equation of surface f(x,y)=12x2+1 and plane is z=0,y=0,y=3x,x=4.

Formula used:

The procedure to calculate volume of surface z=f(x,y),

Step-1: Write the equation of surface in the form z=f(x,y)

Step-2: Sketch the projected region R in the x-y-plane.

Step-3: Determine the order of limits of integration.

Step-4: Evaluate the volume of solid region,

Volume=mnabf(x,y)dxdy

Here, the projected region is R:

myn and axb

Calculation:

Consider equation of surface,

f(x,y)=12x2+1

The region R: z=0,y=0,y=3x,x=4

The following table shown different coordinate of (x,y) for y=3x.

 x-Coordinate y-Coordinate (x,y) Coordinate 0 0 (0,0) 4 12 (4,12)

The graph of region R: z=0,y=0,y=3x,x=4.

For order of limit, the bounds for x are 0x4 and bounds for y are 0y3x.

The volume for the solid region is,

Volume=0403x(12x2+1)dydx

For order of limit the bounds for y are 0y12 and bounds for x are 0xy3.

The volume for the solid region is,

Volume=0120y3(12x2+1)dxdy

Integration is independent from order of limit

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