The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method.

Question

1. y = -x^2 + 6x - 8, y = 0; about the y-axis
2. y^2 - x^2 = 1, y = 2; about the x-axis
3. x^2+(y-1)^2 =1 about the y axis
4. x =(y-1)^2 , x-y=1, about x= -1

Answer 1

1. y = -x^2 + 6x - 8, y = 0; about the y-axis:

To find the volume of the solid generated by rotating the region bounded by the given curves about the y-axis, we can use the disk method. First, we need to solve for x in terms of y:

x^2 = 6x - y - 8

x^2 - 6x + (y + 8) = 0

Now, we can use the quadratic formula to solve for x:

x = (6 ± sqrt(36 - 4(y + 8))) / 2

x = (6 ± sqrt(36 - 4y - 32)) / 2

x = (6 ± sqrt(4 - 4y)) / 2

Since x is non-negative in the region of interest, we take the positive square root:

x = (6 + sqrt(4 - 4y)) / 2

Now, we can find the limits of integration by solving the equation y = -x^2 + 6x - 8 for y = 0:

0 = -x^2 + 6x - 8

x^2 - 6x + 8 = 0

(x - 4)(x - 2) = 0

x = 2, 4

So, the limits of integration are y = -4 and y = 0. The volume V can be found using the disk method:

V = pi * ∫(from -4 to 0) ((6 + sqrt(4 - 4y)) / 2)^2 dy

V = pi * ∫(from -4 to 0) (36 + 12*sqrt(4 - 4y) - 4(4 - 4y)) / 4 dy

V = pi * ∫(from -4 to 0) (9 + 3*sqrt(4 - 4y) - (4 - 4y)) dy

Now, you can evaluate the integral and find the volume.

2. y^2 - x^2 = 1, y = 2; about the x-axis:

This region is symmetric about the x-axis, so we can find the volume by using the washer method. First, let's find the x-intercepts of y^2 - x^2 = 1:

x^2 = y^2 - 1

x = ±sqrt(y^2 - 1)

Now, we can use the washer method to find the volume of the solid generated by rotating the region bounded by the given curves about the x-axis. The volume V can be found using the washer method:

V = pi * ∫(from 1 to 2) (sqrt(y^2 - 1))^2 - (-sqrt(y^2 - 1))^2 dy

Since (sqrt(y^2 - 1))^2 = y^2 - 1 for both positive and negative square roots, the integral becomes:

V = pi * ∫(from 1 to 2) (2y^2 - 2) dy

Now, you can evaluate the integral and find the volume.

3. x^2 + (y - 1)^2 = 1, about the y-axis:

To find the volume of the solid generated by rotating the region bounded by the given curve about the y-axis, we can use the disk method. First, we need to solve for x in terms of y:

x^2 = 1 - (y - 1)^2

x = ±sqrt(1 - (y - 1)^2)

Since x is non-negative in the region of interest, we take the positive square root:

x = sqrt(1 - (y - 1)^2)

Now, we can find the limits of integration by solving the equation x^2 + (y - 1)^2 = 1 for x = 0:

0 + (y - 1)^2 = 1

(y - 1)^2 = 1

y - 1 = ±1

y = 2, 0

So, the limits of integration are y = 0 and y = 2. The volume V can be found using the disk method:

V = pi * ∫(from 0 to 2) (sqrt(1 - (y - 1)^2))^2 dy

V = pi * ∫(from 0 to 2) (1 - (y - 1)^2) dy

Now, you can evaluate the integral and find the volume.

4. x = (y - 1)^2, x - y = 1; about x = -1:

First, we need to find the points of intersection of the two curves:

x = (y - 1)^2

x - y = 1

Substituting x from the first equation into the second equation:

(y - 1)^2 - y = 1

Expanding and simplifying:

y^2 - 2y + 1 - y = 1

y^2 - 3y = 0

y(y - 3) = 0

y = 0, 3

The limits of integration are y = 0 and y = 3. The volume V can be found using the washer method. First, find x in terms of y for the second equation:

x = y + 1

Now, set up the integral:

V = pi * ∫(from 0 to 3) (((y - 1)^2 + 1)^2 - ((y + 1) + 1)^2) dy

V = pi * ∫(from 0 to 3) (((y - 1)^2 + 1)^2 - (y + 2)^2) dy

Now, you can evaluate the integral and find the volume.