   Chapter 7.9, Problem 5E ### 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 = 2 − 6 x − 6 y

To determine

To calculate: The volume of solid region bounded in first octane and the plane z=26x6y.

Explanation

Given information:

The provided equation of plane is z=26x6y.

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=26x6y

The following table shown different coordinate of (x,y,z) for z=26x6y.

 x-Coordinate y-Coordinate z-Coordinate (x,y,z) Coordinate 0 0 2 (0,0,2) 0 6 0 (0,6,0) 6 0 0 (6,0,0)

The graph of function z=26x6y in first octant shown below.

Substitute, 0 for z in equation of plane z=26x6y.

26x6y=06x6y=2y=13x

The following table shown different coordinate of (x,y) for y=13x.

 x-Coordinate y-Coordinate (x,y) Coordinate 13 0 (13,0) 0 13 (0,13)

The graph of equation y=13x 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 0x13 and bounds for y are 0y13x and another choose to set up the double integration is to choose y as the outer variable. with that choice, The bounds for y are 0y13 and bounds for x are 0xy13

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