#### 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= 2π4π3-3

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 solid revolution about x-axis is

V= ∫abA(x)dx

**2) Given:**

The region is bounded by y=1+secx, y=3; about y=1.

**3) Calculation:**

The region is bounded by y=1+secx, y=3 rotated 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 3-1=2 and the inner radius is 1+secx-1=secx

So, its cross sectional area is

Ax=πouter radius2-πinner radius2 =π22-πsecx2

=π4-sec2x

The region of integration is bounded by y=1+secxand y=3

At intersection of these curves

1+secx=3

secx=2

Hence,

x= -π3 or x=π3

The solid lies between

x=-π3 and x= π3

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

V= ∫-π3π3Axdx=∫-π3π3π(4-sec2x)dx

From graph or otherwise,fx=4 –sec2x is an even function (because cos is an even function).

Therefore,

V= 2∫0π3π(4-sec2x)dx

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

V=2π4x-tanx0π3

By substituting the limits of integration,

=2π 4π3-tanπ3-40-tan0

=2π4π3-3-0

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

V=2π 4π3-3

**Conclusion:**

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

V=2π 4π3-3

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

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