Chapter 14.2, Problem 34E

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347

Chapter
Section

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347
Textbook Problem

# Finding Volume In Exercises 29-34, set up and evaluate a double integral to find the volume of the solid bounded by the graphs of the equations. x 2 + z 2 = 1 ,   y 2 + z 2 = 1,           first   octant

To determine

To calculate: The volume of the given solid that is bound by the graphs of the equations x2+z2=1,y2+z2=1 in first octant.

Explanation

Given:

The equation of volume of solid represented by x2+z2=1,y2+z2=1 in first octant.

Formula used:

The volume of the solid is given by,

V=âˆ¬Rf(x,y)dA

Where, f(x,y)=z

Calculation:

Take into consideration the solid bounded by the graphs of the equations.

x2+z2=1,y2+z2=1 in first octant.

The aim is to set up the double integral in order to calculate the volume of the solid.

The intersection of these two graphs is calculated by the steps shown below:

From the two given equations,

1âˆ’x2=1âˆ’y2âˆ’x2=âˆ’y2

x=y (Since graph is in 1st octant)

The limits of x and y are:

R={(x,y)0â‰¤xâ‰¤1,0â‰¤yâ‰¤x}

The graph of the region is shown below,

In the given case, it can be observed from the given graph that there are two regions that are equal. One of them is above the line x=y and the other is below the line x=y.

The volume of the solid bounded y in the region R is given by,

V=âˆ¬Rf(x,y)dA

Let,

z=f(x,y)=1âˆ’x2

Hence, the volume of the solid bounded by the given curves in first octant can be obtained by,

V=âˆ¬Rf(x,y)dA=âˆ¬R1f(x,y)dA+

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