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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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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 = 1 4 x 2 y

To determine

To calculate: The volume of solid region bounded in first octane and the plane z=14x2y.

Explanation

Given information:

The provided equation of plane is z=14x2y.

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=14x2y

The following table shown different coordinate of (x,y,z) for z=14x2y.

x-Coordinate y-Coordinate z-Coordinate (x,y,z) Coordinate
0 0 1 (0,0,2)
0 12 0 (0,12,0)
14 0 0 (14,0,0)

The graph of function z=14x2y in first octant shown below.

Substitute, 0 for z in equation of plane z=14x2y.

14x2y=04x+2y=1y=122x

The following table shown different coordinate of (x,y) for y=122x.

x-Coordinate y-Coordinate (x,y) Coordinate
14 0 (14,0)
0 12 (0,12)

The graph of equation y=122x 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 0x14 and bounds for y are 0y122x 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 0y12 and bounds for x are 0xy214

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