Volume of Solids by Disk and Washer Methods

1) What is the volume formula for a solid with cross-sections perpendicular to the x-axis using the disk method?

V = ∫[a to b] π[f(x)]² dx

V = ∫[a to b] π[f(x) - g(x)]² dx

V = ∫[a to b] A(y) dy

V = ∫[a to b] 2π f(x) dx

The volume formula using the disk method calculates the volume of a solid by integrating the area of circular cross-sections perpendicular to the x-axis. Here, \(f(x)\) represents the radius of each disk at a given x-value. By squaring the radius and multiplying by π, we find the area of each circular cross-section. Integrating this area from \(a\) to \(b\) sums the volumes of all disks, yielding the total volume of the solid. This approach is particularly effective for solids of revolution where the shape is defined by a function above the x-axis.

Explanation

The volume formula using the disk method calculates the volume of a solid by integrating the area of circular cross-sections perpendicular to the x-axis. Here, \(f(x)\) represents the radius of each disk at a given x-value. By squaring the radius and multiplying by π, we find the area of each circular cross-section. Integrating this area from \(a\) to \(b\) sums the volumes of all disks, yielding the total volume of the solid. This approach is particularly effective for solids of revolution where the shape is defined by a function above the x-axis.

2) When revolving the region under y = √x over [1, 4] about the x-axis, what is the resulting volume?

8π

15π/2

7π/6

2π/15

To find the volume of the solid formed by revolving the curve \( y = \sqrt{x} \) from \( x = 1 \) to \( x = 4 \) around the x-axis, we use the disk method. The volume \( V \) is given by the integral \( V = \pi \int_{1}^{4} (\sqrt{x})^2 \, dx \). This simplifies to \( V = \pi \int_{1}^{4} x \, dx \). Evaluating the integral, we find \( V = \pi \left[ \frac{x^2}{2} \right]_{1}^{4} = \pi \left( \frac{16}{2} - \frac{1}{2} \right) = \pi \cdot \frac{15}{2} = \frac{15\pi}{2} \).

Explanation

To find the volume of the solid formed by revolving the curve \( y = \sqrt{x} \) from \( x = 1 \) to \( x = 4 \) around the x-axis, we use the disk method. The volume \( V \) is given by the integral \( V = \pi \int_{1}^{4} (\sqrt{x})^2 \, dx \). This simplifies to \( V = \pi \int_{1}^{4} x \, dx \). Evaluating the integral, we find \( V = \pi \left[ \frac{x^2}{2} \right]_{1}^{4} = \pi \left( \frac{16}{2} - \frac{1}{2} \right) = \pi \cdot \frac{15}{2} = \frac{15\pi}{2} \).

3) In the washer method, the cross-sectional area A is defined as ____.

In the washer method, the cross-sectional area A represents the area of a circular washer formed by two concentric circles, where R is the outer radius and r is the inner radius. The formula A = π(R² - r²) calculates the area by subtracting the area of the smaller circle (πr²) from the area of the larger circle (πR²). This difference gives the area of the annular region, which is essential for determining the volume of solids of revolution when rotating a region around an axis.

Explanation

In the washer method, the cross-sectional area A represents the area of a circular washer formed by two concentric circles, where R is the outer radius and r is the inner radius. The formula A = π(R² - r²) calculates the area by subtracting the area of the smaller circle (πr²) from the area of the larger circle (πR²). This difference gives the area of the annular region, which is essential for determining the volume of solids of revolution when rotating a region around an axis.

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4) When finding the volume of the solid formed by revolving the region enclosed by y = x and y = x² about the x-axis, what are the outer radius R and inner radius r respectively?

R = x², r = x

R = x, r = x²

R = 1, r = x

R = x, r = 1

When revolving the region enclosed by the curves \( y = x \) and \( y = x^2 \) about the x-axis, the outer radius \( R \) corresponds to the upper function, which is \( y = x \), and the inner radius \( r \) corresponds to the lower function, which is \( y = x^2 \). This configuration creates a solid of revolution where the area between these two curves defines the volume. Thus, for the given functions, the outer radius is \( x \) and the inner radius is \( x^2 \).

Explanation

When revolving the region enclosed by the curves \( y = x \) and \( y = x^2 \) about the x-axis, the outer radius \( R \) corresponds to the upper function, which is \( y = x \), and the inner radius \( r \) corresponds to the lower function, which is \( y = x^2 \). This configuration creates a solid of revolution where the area between these two curves defines the volume. Thus, for the given functions, the outer radius is \( x \) and the inner radius is \( x^2 \).

5) The volume of the solid obtained by revolving the region enclosed by y = x and y = x² about the x-axis equals 2π/15.

True

False

To find the volume of the solid formed by revolving the region between the curves \(y = x\) and \(y = x^2\) around the x-axis, we use the disk method. The volume \(V\) is calculated using the integral \(V = \pi \int_{a}^{b} (R^2 - r^2) \, dx\), where \(R\) is the outer radius and \(r\) is the inner radius. The curves intersect at \(x = 0\) and \(x = 1\), and after evaluating the integral from 0 to 1, the result is indeed \( \frac{2\pi}{15} \), confirming the statement as true.

Explanation

To find the volume of the solid formed by revolving the region between the curves \(y = x\) and \(y = x^2\) around the x-axis, we use the disk method. The volume \(V\) is calculated using the integral \(V = \pi \int_{a}^{b} (R^2 - r^2) \, dx\), where \(R\) is the outer radius and \(r\) is the inner radius. The curves intersect at \(x = 0\) and \(x = 1\), and after evaluating the integral from 0 to 1, the result is indeed \( \frac{2\pi}{15} \), confirming the statement as true.

6) When revolving the region enclosed by y = x and y = x² about the line y = 2, what are the outer radius R and inner radius r?

R = 2 - x, r = 2 - x²

R = 2 + x², r = 2 + x

R = 2 - x², r = 2 - x

R = x² - 2, r = x - 2

When revolving the area between the curves y = x and y = x² around the line y = 2, the outer radius (R) is determined by the distance from the line y = 2 to the upper curve y = x², resulting in R = 2 - x². The inner radius (r) is the distance from the line y = 2 to the lower curve y = x, giving r = 2 - x. These radii are essential for calculating the volume of the solid formed by this revolution using the washer method.

Explanation

When revolving the area between the curves y = x and y = x² around the line y = 2, the outer radius (R) is determined by the distance from the line y = 2 to the upper curve y = x², resulting in R = 2 - x². The inner radius (r) is the distance from the line y = 2 to the lower curve y = x, giving r = 2 - x. These radii are essential for calculating the volume of the solid formed by this revolution using the washer method.

7) Match each example with its computed volume.

Region under y = √x over [1,4] revolved about x-axis

Region enclosed by y = x and y = x² revolved about x-axis

Region enclosed by y = x and y = x² revolved about y = 2

Region under y = x³ between y = 0 and y = 8 revolved about y-axis

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8) To find the volume of the solid obtained when the region under y = x³ between y = 0 and y = 8 is revolved about the y-axis, which of the following correctly sets up the integral?

V = ∫[0 to 8] π(y^(1/3))² dy

V = ∫[0 to 8] π(y^(1/2))² dy

V = ∫[0 to 8] π(y³)² dy

V = ∫[0 to 8] π(y^(2/3)) dy

To find the volume of the solid formed by revolving the region under the curve \( y = x^3 \) around the y-axis, we first express \( x \) in terms of \( y \), giving \( x = y^{1/3} \). The volume of revolution can be calculated using the disk method, where the area of the circular cross-section is \( \pi x^2 \). Substituting \( x \) yields \( \pi (y^{1/3})^2 \). Thus, integrating this from \( y = 0 \) to \( y = 8 \) provides the correct setup for the volume integral.

Explanation

To find the volume of the solid formed by revolving the region under the curve \( y = x^3 \) around the y-axis, we first express \( x \) in terms of \( y \), giving \( x = y^{1/3} \). The volume of revolution can be calculated using the disk method, where the area of the circular cross-section is \( \pi x^2 \). Substituting \( x \) yields \( \pi (y^{1/3})^2 \). Thus, integrating this from \( y = 0 \) to \( y = 8 \) provides the correct setup for the volume integral.

Volume of Solids by Disk and Washer Methods

1) What is the volume formula for a solid with cross-sections perpendicular to the x-axis using the disk method?

2)

What first name or nickname would you like us to use?

2) When revolving the region under y = √x over [1, 4] about the x-axis, what is the resulting volume?

3) In the washer method, the cross-sectional area A is defined as ____.

4) When finding the volume of the solid formed by revolving the region enclosed by y = x and y = x² about the x-axis, what are the outer radius R and inner radius r respectively?

5) The volume of the solid obtained by revolving the region enclosed by y = x and y = x² about the x-axis equals 2π/15.

6) When revolving the region enclosed by y = x and y = x² about the line y = 2, what are the outer radius R and inner radius r?

7) Match each example with its computed volume.

8) To find the volume of the solid obtained when the region under y = x³ between y = 0 and y = 8 is revolved about the y-axis, which of the following correctly sets up the integral?