#### 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. To evaluate**:

The integral by using the calculator up to five decimal places

#### Answer

i. V=2π∫00.7861-x2-x4dx

ii. V≈3.54459

#### Explanation

**1) Concept:**

i. If the cross section is a washer with inner radius rin and outer radius rout, then 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 x-axis is

V= ∫abA(x)dx

**2) Given:**

y=x2, x2+y2=1, y≥0, about the x-axis

**3) Calculation:**

Region bounded by the given curves is shown below:

The region rotates about the x-axis, so the cross-section is perpendicular to the x-axis and it forms a washer.

To find the outer and inner radius, solve the given equation for y because limits of integration are in terms of x

x2+y2=1

Subtract x2 from both sides,

y2=1-x2

By taking square root of both sides,

y=1-x2 …………….………….. since y≥0

The outer radius is the distance from the upper part of the circle to the axis of rotation x axis

Outer radius =1-x2-0 = 1-x2

The inner radius is the distance from the curve to the axis of rotation x axis

Inner radius =x2-0=x2

To find limits of integration equate y=x2 and y=1-x2

Therefore,

x2=1-x2

By squaring both sides,

x4=1-x2

x4+ x2-1=0

Substitute, x2=a

a2+a-1=0

By using quadratic formula, value of a is

a=5-12≈0.786

The solid lies between x=0.786 and x= -0.786

By using the concept

V=π∫-aa1-x22-x22dx

V=π∫-aa1-x2-x4dx

Substitute a=0.786 and -a= -0.786, it becomes

V=π∫-0.7860.7861-x2-x4dx

By using symmetry,

V=2π∫00.7861-x2-x4dx

Now use the calculator to evaluate the above integration.

V≈3.54459

Therefore,

V=2π∫00.7861-x2-x4dx

V≈3.54459

**Conclusion:**

V=2π∫00.7861-x2-x4dx

V≈3.54459

#### 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 y-axis

**ii. To evaluate**:

The integral by using the calculator up to five decimal places

#### Answer

i. V= π ∫00.618 y dy+ π ∫0.61811-y2dy

ii. V≈0.99998

#### 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=x2, x2+y2=1, y≥0, about y-axis

**3) Calculation:**

Bounded region of the given curves is shown below:

The region rotates about the y-axis, so the cross-section is perpendicular to the y-axis and it forms adisc. Also, the bounded region can be divided in two parts A1 and A2. So, find volume for both regions separately, and then add to get the total volume.

To find the radius for each part separately, solve the given equations for x since limits of integration are in terms of y

x2+y2=1

Subtract x2 from both sides,

x2=1-y2

By taking square root of both sides,

x= 1-y2 ………….since y≥0

Also solve y=x2 for x

By taking square root of both sides, it becomes

x= y ………….. since y≥0

In region A1, area of cross section is

A1= π 1-y22

In region A1, solid lies between y= 0.618 and y= 1.

By using the concept

V1=∫0.6181π 1-y22dy

……………………………….(1)

In region A2, area of cross section is

A2= π y2

In region A2, solid lies between y= 0 and y= 0.618.

By using concept,

V2= ∫00.618π y2dy

……………………………….(1)

By using (1) and (2),

V= ∫00.618π y2dy+ ∫0.6181π 1-y22dy

V= π ∫00.618 y dy+ π ∫0.61811-y2dy

By using the calculator

V 0.99998

Therefore,

V= π ∫00.618 y dy+ π ∫0.61811-y2dy

V 0.99998

**Conclusion:**

V= π ∫00.618 y dy+ π ∫0.61811-y2dy

V 0.99998