#### To determine

**To find:**

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

#### Answer

V=29π30

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 y-axis is

V= ∫abA(y)dy

**2) Given:**

y=x2,x=y2,x=1;the line x=-1

**3) Calculation:**

The region bounded by y=x2,x=y2,x=1 rotated about the line x=-1 are shown below:

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

A cross section of the solid is the washer with the outer radius is y--1= y+1 (distance from x=-1 to x=y) and the inner radius is y2--1=y2+1 (distance from x=-1 to x=y2).

So, its cross sectional area is

Ay=πouter radius2-πinner radius2 =πy+12-πy2+12

=πy+2y+1-y4+2y2+1

A(x)=πy+2y-y4-2y2

The region of integration is bounded by y=x2 and x=y2

So at point of intersection,

y4=y

y4-y=0

yy3-1=0

y=0 and y3=1

The solid lies between y=0 and y= 1

Therefore, the volume of the solid is

V= ∫01Aydy=∫01πy+2y12-y4-2y2dy

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

V=πy22+2y3232-15y5-2y3301

=πy22+4y323-15y5-2y3301

By substituting the limits of integration,

=π (1)22+4(1)323-(1)55-2(1)33-(0)22+4(0)323-055-2033

=π 12+43-15-23-0

=π1530+4030-630-2030

V=π15+40-6-2030

V=29π30

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

V=29π30

**Conclusion:**

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

V=29π30

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