   Chapter 7.9, Problem 4E ### 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 1-6, find the volume of the solid region bounded in the first octant by the plane. See Example 1. z = 16 − 8 x − 4 y

To determine

To calculate: The volume of solid region bounded in first octane and the plane z=168x4y.

Explanation

Given information:

The provided equation of plane is z=168x4y.

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 plane,

z=168x4y

The following table shown different coordinate of (x,y,z) for z=168x4y.

 x-Coordinate y-Coordinate z-Coordinate (x,y,z) Coordinate 0 0 16 (0,0,16) 0 4 0 (0,4,0) 2 0 0 (2,0,0)

The graph of function z=168x4y in first octant shown below.

Substitute, 0 for z in equation of plane z=168x4y.

168x4y=08x+4y=16y=42x

The following table shown different coordinate of (x,y) for y=42x.

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

The graph of equation y=42x is

One way to set up the double integration is to choose x as the outer variable with that choice. The bounds for x are 0x2 and bounds for y are 0y42x and another choose to set up the double integration is to choose y as the outer variable

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