Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

Question

1. y =x + 1, y = 0, r = 0, x = 2; about the x-axis
2. y = square root x - 1, y = 0, x = 5; about the x-axis
3. x=2 square root y , x=0, y=9; about the y-axis
4. y=x^3 , y=x, x>0; about the x-axis
5. y^2 = x, x = 2y; about the y-axis
6. y = x^2,x=y^2, about y=1
7. y= 1 +sec x, y= 3 , about y=1
8. y= x^3, y=0 about x=2
9. x=y^2, x=1-y^2 about x =3

Answer 1

1. For the first problem, we need to find the volume of the solid obtained by rotating the region bounded by the curves y = x + 1, y = 0, r = 0, and x = 2 about the x-axis.

First, let's sketch the region and the solid:

To find the volume of the solid, we can use the formula for the volume of a solid of revolution:

V = ∫[a, b] π(R(x)^2 - r(x)^2) dx

where R(x) is the distance from the x-axis to the outer curve and r(x) is the distance from the x-axis to the inner curve.

In this case, R(x) = x + 1 and r(x) = 0, so we have:

V = ∫[0, 2] π(x + 1)^2 dx

= π ∫[0, 2] (x^2 + 2x + 1) dx

= π [x^3/3 + x^2 + x] [0, 2]

= π (8/3 + 4 + 2)

= 14π/3

Therefore, the volume of the solid is 14π/3.

2. For the second problem, we need to find the volume of the solid obtained by rotating the region bounded by the curves y = √x - 1, y = 0, and x = 5 about the x-axis.

Let's first sketch the region and the solid:

To find the volume of the solid, we can again use the formula for the volume of a solid of revolution:

V = ∫[a, b] π(R(x)^2 - r(x)^2) dx

In this case, R(x) = √x - 1 and r(x) = 0, so we have:

V = ∫[0, 5] π(√x - 1)^2 dx

= π ∫[0, 5] (x - 2√x + 1) dx

= π [x^2/2 - 4x^(3/2)/3 + x] [0, 5]

= 29π/3

Therefore, the volume of the solid is 29π/3.

3. For the third problem, we need to find the volume of the solid obtained by rotating the region bounded by the curves x = 2√y, x = 0, and y = 9 about the y-axis.

Let's first sketch the region and the solid:

To find the volume of the solid, we can again use the formula for the volume of a solid of revolution:

V = ∫[a, b] π(R(y)^2 - r(y)^2) dy

In this case, R(y) = 2√y and r(y) = 0, so we have:

V = ∫[0, 9] π(2√y)^2 dy

= 4π ∫[0, 9] y dy

= 4π [y^2/2] [0, 9]

= 162π

Therefore, the volume of the solid is 162π.

4. For the fourth problem, we need to find the volume of the solid obtained by rotating the region bounded by the curves y = x^3, y = x, and x > 0 about the x-axis.

Let's first sketch the region and the solid:

To find the volume of the solid, we can again use the formula for the volume of a solid of revolution:

V = ∫[a, b] π(R(x)^2 - r(x)^2) dx

In this case, R(x) = x^3 and r(x) = x, so we have:

V = ∫[0, 1] π(x^6 - x^2) dx

= π [x^7/7 - x^3/3] [0, 1]

= 2π/21

Therefore, the volume of the solid is 2π/21.

5. For the fifth problem, we need to find the volume of the solid obtained by rotating the region bounded by the curves y^2 = x and x = 2y about the y-axis.

Let's first sketch the region and the solid:

To find the volume of the solid, we can again use the formula for the volume of a solid of revolution:

V = ∫[a, b] π(R(y)^2 - r(y)^2) dy

In this case, R(y) = 2y and r(y) = √y, so we have:

V = ∫[0, 4] π(4y^2 - y) dy

= π [4y^3/3 - y^2/2] [0, 4]

= 32π/3

Therefore, the volume of the solid is 32π/3.

6. For the sixth problem, we need to find the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and x = y^2 about the line y = 1.

Let's first sketch the region and the solid:

To find the volume of the solid, we can use the formula for the volume of a solid of revolution with respect to a line other than the x- or y-axis:

V = ∫[a, b] 2π(x - h)(R(x) - h) dx

where h is the distance from the axis of revolution to the line y = 1, which in this case is 1.

R(x) is the distance from the axis of revolution to the outer curve, which is y = x^2. Since we are rotating about the line y = 1, we need to shift the curve up by 1, so R(x) = 1 + x^2 - 1 = x^2.

We can find the inner curve by solving x = y^2 for y, which gives y = ±√x. We only need the curve that is above y = 1, so we have y = √x. The distance from the axis of revolution to this curve is R(x) - h = √x - 1.

Substituting these values into the formula, we have:

V = ∫[0, 1] 2π(x - 1)(x^2 - 1) dx

= 2π ∫[0, 1] (x^3 - x^2 - x + 1) dx

= 2π [x^4/4 - x^3/3 - x^2/2 + x] [0, 1]

= 11π/6

Therefore, the volume of the solid is 11π/6.

7. For the seventh problem, we need to find the volume of the solid obtained by rotating the region bounded by the curves y = 1 + sec x, y = 3, and the lines x = 0 and x = π/3 about the line y = 1.

Let's first sketch the region and the solid:

To find the volume of the solid, we can again use the formula for the volume of a solid of revolution with respect to a line other than the x- or y-axis:

V = ∫[a, b] 2π(x - h)(R(x) - h) dx

where h is the distance from the axis of revolution to the line y = 1, which in this case is 1.

R(x) is the distance from the axis of revolution to the outer curve, which is y = 3. Since we are rotating about the line y = 1, we need to shift the curve down by 1, so R(x) = 2.

We can find the inner curve by solving y = 1 + sec x for x. This gives sec x = y - 1, so x = sec^-1(y - 1). The distance from the axis of revolution to this curve is R(x) - h = 1 + sec^-1(y - 1) - 1 = sec^-1(y - 1).

Substituting these values into the formula, we have:

V = ∫[1, 3] 2π(sec^-1(y - 1))(2 - 1) dy

= π ∫[1, 3] (sec^-1(y - 1)) dy

= π [(y - 1)sec^-1(y - 1) + √(y - 1)(y - 2) + π/2 - 1] [1, 3]

= π [(2sec^-1 2 + √2 + π/2 - 1) - (√2/2 + π/6 - 1)]

= 2π(sec^-1 2 + √2/2 - π/6)

Therefore, the volume of the solid is 2π(sec^-1 2 + √2/2 - π/6).

8. For the eighth problem, we need to find the volume of the solid obtained by rotating the region bounded by the curves y = x^3 and y = 0 about the line x = 2.

Let's first sketch the region and the solid:

To find the volume of the solid, we can use the formula for the volume of a solid of revolution with respect to the line x = 2:

V = ∫[a, b] 2π(y - k)(R(y) - k) dy

where k is the distance from the axis of revolution to the line x = 2, which in this case is 2.

R(y) is the distance from the axis of revolution to the outer curve, which is y = x^3. Since we are rotating about the line x = 2, we need to shift the curve to the right by 2, so R(y) = (y^(1/3) + 2).

The inner curve is the x-axis, which is y = 0. The distance from the axis of revolution to this curve is R(y) - k = y^(1/3).

Substituting these values into the formula, we have:

V = ∫[0, 8] 2π(y - 2)(y^(1/3) + 2 - 2) dy

= 2π ∫[0, 8] y^(4/3) - 4y^(1/3) + 4 dy

= 2π [(3y^(7/3))/7 - 6y^(4/3)/4 + 4y] [0, 8]

= 384π/7

Therefore, the volume of the solid is 384π/7.

9. For the ninth problem, we need to find the volume of the solid obtained by rotating the region bounded by the curves x = y^2 and x = 1 - y^2 about the line x = 3.

Let's first sketch the region and the solid:

To find the volume of the solid, we can use the formula for the volume of a solid of revolution with respect to the line x = 3:

V = ∫[a, b] 2π(x - h)(R(x) - h) dx

where h is the distance from the axis of revolution to the line x = 3, which in this case is 3.

R(x) is the distance from the axis of revolution to the outer curve, which is x = 1 - y^2. Since we are rotating about the line x = 3, we need to shift the curve to the right by 3, so R(x) = 1 - y^2 + 3 = 4 - y^2.

We can find the inner curve by solving x = y^2 for y. This gives y = ±√x. We only need the curve that is to the right of x = 3, so we have y = √x. The distance from the axis of revolution to this curve is R(x) - h = (4 - x)^(1/2).

Substituting these values into the formula, we have:

V = ∫[0, 1] 2π(x - 3)(4 - x)^(1/2) dx

= 4π ∫[0, 1] (x - 3)(4 - x)^(1/2) dx

= 4π [(-15/8)(4 - x)^(3/2) + (5/2)(4 - x)^(3/2) + (16/3)(4 - x)^(3/2)] [0, 1]

= (17π)/3

Therefore, the volume of the solid is (17π)/3.