#### 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 line y=2.

**ii. To evaluate**:

The integral by using the calculator up to five decimal places

#### Answer

i. V=16π∫021-x24dx

ii. V≈78.95684

#### 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 a solid revolution about the x-axis is

V= ∫abA(x)dx

**2) Given:**

x2+4y2=4, about y=2

**3) Calculation:**

Bounded region of the given curves is shown below:

The region rotates about the line =2, 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 x2+4y2=4 for y since the limits of integration are in terms of x

x2+4y2=4

Divide both sides by 4

x24+y2=1

Subtract x24 from both sides

y2=1-x24

By taking square root of both sides

y=±1-x24

So the outer radius is the distance of the bottom half of the ellipse to the axis of rotation y=2

Outer radius=2--1-x24=2+1-x24

The inner radius is the distance of the top half of the ellipse to the axis of rotation y=2

Inner radius =2-1-x24

The solid lies between x=-2 and x=2.

By using the concept

V=π∫-222+1-x242-2-1-x242dx

By using symmetry and expanding square terms

=2π∫024+41-x24+1-x24-4-41-x24+ 1-x24dx

=2π∫0281-x24dx

V=16π∫021-x24dx

By using the calculator (refer to exercise)

V≈78.95684

Therefore,

V=16π∫021-x24dx

V≈78.95684

**Conclusion:**

V=16π∫021-x24dx

V≈78.95684

#### 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 x=2.

**ii. To evaluate**:

The integral by using the calculator up to five decimal places

#### Answer

i. V=32π∫011-y2dy

ii. V≈78.95684

#### 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 y-axis is

V= ∫abA(y)dy

**2) Given:**

x2+4y2=4, about x=2

**3) Calculation:**

Bounded region of the given curves is shown below:

The region rotates about the line =2, so the cross-section is perpendicular to the y-axis, and it forms a washer.

To find the outer and inner radius, solve the given equation x2+4y2=4 for x since limits of integration are in terms of y.

Solve the given equation for x.

x2+4y2=4

Subtract 4y2 from both sides

x2=4-4y2

Factor out 4 from the left side of the equation then take square root of both sides

x=±21-y2

So the outer radius is the distance of the left half of the ellipse to the axis of rotation x=2

Outer radius=2--21-y2

Outer radius=2+21-y2

The inner radius is the distance of the right half of the ellipse to the axis of rotation x=2

Inner radius=2-21-y2

The solid lies between y=-1 and y=1.

By using the concept

V=π∫-112+21-y22-2-21-y22dy

=π∫-114+81-y2+41-y2-4-81-y2+4(1-y2)dy

=π∫-11161-y2dy

By using symmetry

=2π∫01161-y2dy

=32π∫011-y2dy

By using the calculator

V≈78.95684

Therefore,

V=32π∫011-y2dy

V≈78.95684

Therefore,

V=32π∫011-y2dy

V≈78.95684

**Conclusion:**

V=32π∫011-y2dy

V≈78.95684