Area, Volume, and Arc Length Exercises

Answer

The region R looks like

(a) When we rotate R around the x-axis we get the solid

Slicing perpendicular to the x-axis gives us disks of radius y = ln x.

The volume of the disk at position x is

π (ln x)² Δ x.

Since x goes from 1 to 2 in the region R, the volume of the entire solid is

(b) Rotating R around the line x = 2 produces the solid

Slices perpendicular to the line x = 2 are horizontal disks. Since we're working with horizontal disks, we'll have to integrate with respect to y. Solving y = ln x for x gives us x = e^y_.

Looking at the disk at height y, we see that e^y + r must equal 2, so the radius of the disk is

r = 2 – e^y.

The volume of the disk is

π(2 – e^y)² Δ y.

The variable y ranges from 0 to ln 2 in the region R, so the volume of the entire solid is

Answer

Here's the region R:

(a) When R is rotated around the y-axis it gives us the solid

Since R lies along the y-axis for the whole height of R, slices perpendicular to the y-axis are disks.

The disk at height y has radius x = y^1/3 and volume

π(y^1/3)² Δ y = π y^2/3 Δ y.

Since y goes from 0 to 8 in the region R, the volume of the entire solid is

(b) When R is rotated around the line y = 8 we get a solid volume

R lies along the line y = 8 for the entire x-interval [0,2], so slices perpendicular to y = 8 are disks. The radius of the disk at position x is

8 – y = 8 – x³.

The volume of this disk is

π(8 – x³)² Δ x.

Since x goes from 0 to 2 in the region R, the volume of the solid is

When the region doesn't lie along the axis of rotation, a slice perpendicular to the axis of rotation usually looks like a washer, which is just a disk with a hole in the middle. Lucky for us, a frisbee with a hole in it doesn't fly nearly as far. Our faces should be safe from lofted washers.

The volume of a washer is the area of a side multiplied by the thickness. A washer has an outer radius and an inner radius.

To find the area of the side we find the area using the outer radius and subtract out the area of the hole.

Since the area using the outer radius is

π (r_outer)²

and the area of the hole is

π (r_inner)²,

we get a simple equation for the area of the side of the washer:

π[(r_outer)² – (r_inner)²].

Then we multiply by whatever thickness is appropriate (Δ x or Δ y) to find the volume of the washer.

Answer

The region R looks like this:

(a) When we rotate the region R around the line y = 0 we get the solid

We can see from the picture that the outer radius is the distance from the x-axis to the graph y = x² + 1, and the inner radius is the distance from the x-axis to the graph y = 1 – x.

So

r_outer = x² + 1

r_inner = 1 – x.

Since the thickness of a disk is dx, the volume of the disk is

π[(x² + 1)² – (1 – x)² ] Δ x.

There are disks from x = 0 to x = 1. The volume of the solid is

(b) When we rotate R around the line y = 2 we get

The outer radius is the distance from the graph y = 1 – x to the axis of rotation y = 2.

This distance is

r_outer = 2 – (1 – x) = 1 + x

The inner radius is the distance from the graph y = x² + 1 to the axis of rotation y = 2

This distance is

r_inner = 2 – (x² + 1) = 1 – x²

This means the volume of a disk is

π[(r_outer)² – (r_inner)²] Δ x = π[(1 + x)² – (1 – x²)² ] Δ x

Put the pieces together, we get the volume of the solid:

(c) When R is rotated around the line y = -2 we get the solid

The outer radius is the distance from the axis of rotation y = -2 to the graph y = x² + 1.

This distance is

r_outer = (x² + 1) + 2 = x² + 3.

The inner radius is the distance from the axis of rotation y = -2 to the graph y = 1 – x.

This distance is

r_inner = (1 – x) + 2 = 3 – x.

The volume of a disk is

π[(r_outer)² – (r_inner)²] Δ x = π[(x² + 3)² – (3 – x)² ] Δ x,

and the volume of the entire solid is

(d) Rotating R around the line y = 5 produces a solid with a very large hole in the middle.

Let's see if we can do this one without a picture.

The outer radius is the distance from the axis of rotation y = 5 to the graph y = 1 – x.

This is

r_outer = 5 – (1 – x) = 4 + x.

The inner radius is the distance from the axis of rotation y = 5 to the graph y = x² + 1.

This is

r_inner = 5 – (x² + 1) = 4 – x².

The volume of a disk is

π[(r_outer)² – (r_inner)²] Δ x = π[(4 + x)² – (4 – x²)² ] Δ x.

and the volume of the entire solid is

Answer

The region R looks like this:

The variables x and y both go from 0 to 1 in this region, so the limits of integration will be 0 and 1 regardless of whether we're integrating with respect to x or y. That's handy.

(a) Rotating R around the line y = 0 gets us this solid:

Slices perpendicular to the axis of rotation are disks (no hole) with thickness dx and radius

The volume of a disk is

and the volume of the solid is

(b) Rotating R around the line y = 2 produces this solid:

Slices perpendicular to the axis of rotation are washers with thickness dx. The outer radius is constant: r_outer = 2.

The inner radius is the distance from the axis of rotation y = 2 to the line .

We get

The volume of a disk is

and the volume of the solid is

(c) Rotating R around the line x = 1 gets us this solid:

The outer radius is constant: r_outer = 1.

The inner radius is the distance from the axis of rotation to the graph .

So

The volume of a disk is

The variable y goes from 0 to 1, so the volume of the solid is

(d) Rotating R around the line x = -1 produces this solid:

We can see that the inner radius is constant, and the outer radius is the distance from the axis of rotation to the curve .

So

r_inner = 1.

The volume of a disk is

The volume of the solid is