   Chapter 7.9, Problem 23E ### 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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# Finding the Volume of a Solid Region In Exercises 21-24, use a double integral to find the volume of the solid region bounded by the graphs of the equations. See Example 3. z   =   9 − x 2 , z   =   0 , y   =   x   +   2 ,   y   =   0 ,   x   =   0 ,   x   =   2

To determine

To calculate: The volume of solid region bounded by the graph of the equations. z=9x2, z=0, y=x+2, y=0, x=0, x=2

Explanation

Given information:

The provided equation is

z=9x2, z=0, y=x+2, y=0, x=0, x=2

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 solid region,

z=9x2, z=0, y=x+2, y=0, x=0, x=2

The equation of upper solid region is

z=9x2

The region R is bounded by y=x+2, y=0, x=0, x=2,

The following table shown different coordinate of (x,y) for y=x+2.

 x-Coordinate y-Coordinate (x,y) Coordinate 0 2 (0,2) 2 4 (2,4)

The bounds for x are 0x2 and bounds for y are 0yx+2.

The volume for the solid region is,

Volume=020x+2(9x2)dydx

Now integrate with respect to x and apply the limit

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