#### To determine

**To find:**

The volume of the solid obtained by rotating the region bounded by the given curves about the line y=1 and sketch the region, the solid, and a typical disk or washer.

#### Answer

V=11π30

The sketch of the bounded region, the solid, and a typical washer:

#### Explanation

**1) Concept:**

i. If the cross section is a washer with the inner radius rin and the outer radius rout, then the area of the washer is obtained by subtracting the area of the inner disk from the area of the outer disk.

A=π outer radius2 -π inner radius2

ii. The volume of the solid revolution about the x-axis is

V= ∫abA(x)dx

**2) Given:**

y=x2, x=y2; about y=1

**3) Calculation:**

The region bounded by y=x2, x=y2 and solid obtained by rotating the region about the line y=1 is shown below.

Here, the region is rotated about the line y=1, so the cross-section is perpendicular to the x-axis.

A cross section of the solid is the washer with the outer radius 1-x2 and the inner radius 1-x,

So, its cross sectional area is

Ax=πouter radius2-πinner radius2 =π1-x22-π1-x2

=π1-x22-1-x2

=π1-2x2+x4-1-2x+x

Ax=πx4-2x2-x+2x

The region is bounded by y=x2 and x=y2.

At intersection of the curves,

x4=x

x4-x=0

xx3-1=0

x=0 and x=1

The solid lies between x=0 and x= 1

Therefore, the volume of the solid of revolution about line y=1 is

V= ∫01Axdx=∫01π(x4-2x2-x+2x)dx

By using the fundamental theorem of calculus part 2 and the power rule of integration,

V=πx55-2x33-x22+2x323201

=πx55-2x33-x22+4x32301

By substituting the limits of integration,

=π (1)55-2(1)33-(1)22+4(1)323-055-2(0)33-(0)22+4(0)323

=π15-23-12+43-0

=π 630-2030-1530+4030=π6-20-15+4030

V=11π30

Therefore, the volume of the solid obtained by rotating the region bounded by the given curves about the line y=1 is

V=11π30

**Conclusion:**

The volume of the solid obtained by rotating the region bounded by the given curves about the line y=1 is

V=11π30

The sketch of the bounded region, the solid, and a typical washer: