#### 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

V= 763π

Sketch of bounded region, the solid and a typical washer:

#### Explanation

**1) Concept:**

i. If the cross section is a washer with inner radius rin and outer radius rout, then area of 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:**

y=x,y=0, x=2, x=4;about x=1

**3) Calculation:**

The region is bounded by y=x,y=0, x=2, x=4 and rotated about the line x=1 as shown below:

From above figure, for 0≤y≤2, a cross section of the solid is an annulus with outer radius 4-1=3 and inner radius 2-1=1

So it’scross sectional area is,

A1y=πouter radius2-πinner radius2 =π32-π12

=π9-1

A1y=8π

Similarly, for 2≤y≤4, a cross section of the solid is an annulus with outer radius is 4-1=3 and inner radius is y-1.

So its area is given by,

A2y=πouter radius2-πinner radius2 =π32-πy-12

=π9-(y2-2y+1)

A2y=(8+2y-y2)π

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

V= ∫04Aydy=∫02A1ydy+∫24A2ydy

V= ∫028πdy+∫24(8+2y-y2)πdy

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

V=8πy02+π8y+y2-y3324

By substituting limits of integration,

V=8π2-0+π84+42-433-82+22-233

V=16π+π32+16-643-16+4-83

V=16π+π 28-563=44π-563π=132π-56π3

V=76π3

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

V=76π3

**Conclusion:**

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

V=76π3

Sketch of bounded region, the solid and a typical washer: