#### To determine

**To find:**

The volume of the solid obtained by rotating the region bounded by the given curves about x=3 and sketch the region, the solid, and a typical disk or washer.

#### Answer

V=1023π

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 y-axis is

V= ∫abA(y)dy

**2) Given:**

x=y2,x=1-y2; about the line x=3

**3) Calculation:**

The region bounded by x=y2,x=1-y2 and rotated about the line x=3 are shown below.

Here, the region is rotated about the line x=3, so the cross-section is perpendicular to the y-axis.

A cross section of the solid is the washer with the outer radius 3-y2 (distance from x=3 to x=y2) and the inner radius 3-1-y2=2+y2 (distance from x=3 to x=1-y2).

So, its cross sectional area is

Ay=πouter radius2-πinner radius2 =π3-y22-π2+y22

=π9-6y2+y4-4+4y2+y4

Ax=π5-10y2=5π(1-2y2)

The region of integration is bounded by x=y2 and x=1-y2

At intersection,

y2=1-y2

2y2=1

y2=12

y=±12

So, the limits of integrations are

y=-12 to y=12

Therefore, the volume of the solid V= ∫-1212Aydy=∫-12125π(1-2y2)dy

By using the symmetry of curves,

=2∫0125π(1-2y2)dy

V=10π∫012(1-2y2)dy

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

V=10πy-2y33012

By substituting the limits of integration,

=10π 12-21233-0-2(0)33

=10π 12-23·22-0

=10π12-132

V=10π3-132=10π232

V=1023π

Therefore, the volume of the solid obtained by rotating the region bounded by the given curves about line x=3 is

V=1023π

**Conclusion:**

The volume of the solid obtained by rotating the region bounded by the given curves about the line x=3 is

V=1023π

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