#### To determine

**To find:**

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

#### Answer

16565π

#### Explanation

**1) Concept:**

i. If the cross section is a washer with the inner radius rin and 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 solid revolution about y-axis

V= ∫abA(y)dy

**2) Given:**

The region is bounded by x=0,x=9-y2 rotated about the line x=-1

**3) Calculation:**

As the region is bounded by x=0,x=9-y2 rotated about the line x=-1,

From the figure, as the region rotates about the line =-1, the strip is perpendicular to y-axis.

A cross section of the solid is the washer with outer radius is 9-y2--1= 10-y2 (distance from x=-1 to x=9-y2) and inner radius is 0--1=1, which is the distance from x=-1 to x=0.

So its area is given by

Ay=πouter radius2-πinner radius2

=π10-y22-π12

=π100-20y2+y4-1

A(x)=π99-20y2+y4

The region of integration is bounded by x=0,x=9-y2

Therefore, 9-y2=0, y2=9

y=±3

Here the limits of integration are y=-3 to y=3

Therefore, the volume of the solid revolution about line x=-1,

V= ∫-33Aydy

=∫-33π99-20y2+y4dy

As the graph of the function is symmetric about x- axis,fx= 99-20y2+y4 is an even function.

Therefore, V=2π∫0399-20y2+y4dy

By using fundamental theorem of calculus and power rule of integration,

V=2π99y-20y33+15y503

By substituting limits of integration,

=2π 99(3)-20333+1535-99(0)-20033+1505

=2π 297-180+2435-0

=2π117+2435

V=2π585+2435=2π8285

V=16565π

**Conclusion:**

The volume of the solid obtained by rotating the region bounded by the given curves

V=16565π