#### To determine

**Part (a):**

i. **To set up:**

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

**ii. $2**:

The integral by using the calculator up to five decimal places.

#### Answer

i. V=∫0π4π (tanx)2dx

ii. V≈0.67419

#### 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=tanx, y=0, x=π4, about the x-axis

**3) Calculation:**

Bounded region of the given curves is as shown below:

From the above graph

The solid lies between x=0 and x=π4

The region rotates about the x-axis, so the cross-section is perpendicular to the x-axis, and it is a disc with radius tanx

So it’s cross sectional area is Ax=πtan2x

By using the concept

V=∫0π4πtan2xdx

Evaluate the integral by using a calculator Ti-84 plus.

Use the below steps:

1) Click on “Math” option.

2) After that under “Math” search for the option “fnInt(“

3) Write the lower limit, upper limit and function using the proper brackets and required variable.

4) Press enter key; it will display the final answer.

V ≈0.67419

**Conclusion:**

i. V=∫0π4π (tanx)2dx

ii. V≈0.67419

#### To determine

**Part (b):**

**i. To set up:**

An integral for the volume of 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=∫0π4πtan2x+2tanxdx

ii. V≈2.85178

#### 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 y-axis is V= ∫abA(y)dy

**2) Given:**

y=tanx, y=0, x=π4, about y=-1

**3) Calculation:**

Bounded region of the given curves is as shown below:

In the diagram, the cross section is perpendicular to the x-axis and it is a washer.

Thus, the outer radius is the distance from the curve y=tanx to the axis of rotation y= -1 and the inner radius is the distance from the line y=0 to the axis of rotation y= -1.

This gives the inner radius = 0--1=1 and outer radius is tanx-(-1)=tanx+1

So it’s cross sectional area is

Ax=πtanx+12-π12=πtan2x+2tanx

The solid lies between x=0 and x=π4.

By using the concept

V=∫0π4πtan2x+2tanxdx

V=π∫0π4tan2x+2tanxdx

Evaluate the integral by using a calculator Ti-84 plus.

Use the below steps:

1) Click on “Math” option.

2) After that under “Math” search for the option “fnInt(“

3) Write the lower limit, upper limit and function using the proper brackets and required variable.

4) Press enter key; it will display the final answer.

V ≈2.85178

**Conclusion:**

i. V=∫0π4πtan2x+2tanxdx

ii. V≈2.85178