Q. 25

Question

Write the volume of the two solids of revolution that follow in terms of definite integrals that represent accumulations of disks and/or washers. Do not compute the integrals. 

Step-by-Step Solution

Verified
Answer

The required volume is π-3022dy+π03(22-f-1(y)2)dy

1Step 1. Given Information

The given figure is 

2Step 2. Explanation

A solid of revolution is being formed by rotating the region around y-axis.

The given figure can be shown as below,

The region required to set up the integral should be the region bound by the graph and y-axis.

For the part of region below x-axis, it is clear that the region is bound by the function for vertical line at x=2 and y-axis, i.e., x=0. the interval for the y-variable is [-3,0]. Hence, it forms a disk of radius x-units.

Thus, the integral formed by rotation this part of region around y-axis is created as π-3022dy.

Now, the part above x-axis is not bounded by y-axis. But it can be expressed as a washer with outer radius being the graph of x=2 and inner radius is the graph of the function.

3Step 3. Explanation

The equation of graph of the function y=f(x) can be rewritten as x=f-1(y). The y-interval for this region is [0,3]

Thus, the integral of the volume formed by rotating the upper part of given region is expressed as π0322dy-π03(f-1(y))2dy

Add both the integrals to form the integral for complete volume of solid of revolution for entire region.

π-3022dy+π0322dy-π03(f-1(y))2dy=π-3022dy+π03(22-(f-1(y))2)dy