#### To determine

**Part (a):**

i. **To set up:**

An integral for the volume of the solid obtained by rotating the region bounded by the given curves about the x axis.

**ii. $2**:

The integral by usingthe calculator up to five decimal places

#### Answer

i. V=2π∫0π2cos4x dx

ii. V≈3.70110

#### Explanation

**1) Concept:**

i. If the cross section is a disc and the radius of the disc is in terms of x or y then area A=π radius2

ii. The volume of a solid revolution about the x-axis is

V= ∫abA(x)dx

**2) Given:**

y=0, y=cos2x, -π2≤x≤π2, about the x-axis

**3) Calculation:**

Bounded region of the given curves is as shown below:

In the figure, the region is rotated about the x-axis, so the cross-section is perpendicular to the x-axis, and it forms a disk.

So, its radius is cos2x, and it’s cross sectional area is Ax=πcos4x

The solid lies between x=-π2 and x=π2.

By using concept,

V=∫-π2π2πcos2x2dx

V=π∫-π2π2cos2x2dx

By using symmetry of cos2x, integral becomes

V=2π∫0π2cos4x dx

Now use the calculator(refer to exercise 31) to evaluate the above integral. Then we have

V≈3.70110

Therefore,

V=2π∫0π2cos4x dx

and the volume of the solid is V≈3.70110

**Conclusion:**

V=2π∫0π2cos4x dx

and the volume of the solid is V≈3.70110

#### To determine

**Part (b):**

**i. To set up:**

An integral for the volume of the solid obtained by rotating the region bounded by the given curves about the line y=1

**ii. To evaluate**:

The integral by using the calculator up to five decimal places

#### Answer

i. V=2π∫0π22cos2x-cos4xdx

ii. V≈6.16850

#### Explanation

**1) Concept:**

i. If the cross section is a disc and the radius of the disc is in terms of x or y then area A=π radius2

ii. The volume of solid revolution about the y-axis is V= ∫abA(y)dy

**2) Given:**

y=0, y=cos2x, -π2≤x≤π2, about y=1

**3) Calculation:**

Bounded region of the given curves is as shown below:

In the diagram, the region rotates about the line y=1, so the cross-section isperpendicular to the x-axis.

A cross section of the solid is a washer with outer radius 1, which is distance from y=0 to the axis of rotation =1, and inner radius is 1-cos2x which is the distance from the curve y= cos2x to the axis of rotation y=1.

So it’s cross sectional area is

Ax=π(12-1-cos2x2)

By using the concept

The volume of the solid is

V=π∫-π2π212-1-cos2x2dx

By using the symmetry of 1-cos2x

V=2π∫0π212-1-cos2x2dx

V=2π∫0π22cos2x-cos4xdx

By using the calculator,

V≈6.16850

**Conclusion:**

V=2π∫0π22cos2x-cos4xdx

V≈6.16850