Introduction to Volume & Disk Method: Videos & Practice Problems

To find the volume of a three-dimensional solid, we can use integration by considering the area of cross sections. For solids of revolution, formed by rotating a function around an axis, the cross sections are circular, leading to the volume integral $V^2=\int {\pi }_r{r}^{2}dx$ or $\int {\pi }_r{r}^{2}dy$ depending on the axis of rotation. Understanding the radius function and setting appropriate bounds is crucial for accurate volume calculation.

Introduction to Cross Sections

In this section, we explore how to find the volume of a three-dimensional solid using integrals, building on the concept of calculating area under a curve. Instead of summing up rectangles as in two dimensions, we will slice the solid into thin sections to analyze its volume. The example focuses on a solid with a base defined by the function \( f(x) = 4 - x^2 \) over the interval from 0 to 2, where the solid has square cross sections.

To visualize the solid, imagine tilting the two-dimensional \( xy \)-plane backward and extending the area upward, creating a three-dimensional shape. Each slice of this solid resembles a prism with a square cross section. The volume of a single slice can be determined by calculating the area of the cross section and multiplying it by the width of the slice, which is represented as \( dx \).

Since the cross sections are squares, the area \( A(x) \) of one slice can be expressed as the square of the side length. Given that the side length corresponds to the function \( f(x) \), we have:

\[ A(x) = f(x) \times f(x) = (4 - x^2)^2 \]

To find the volume \( V \) of the entire solid, we integrate the area function over the specified interval:

\[ V = \int_{0}^{2} A(x) \, dx = \int_{0}^{2} (4 - x^2)^2 \, dx \]

This integral accounts for all the thin slices from \( x = 0 \) to \( x = 2 \), providing the total volume of the solid. It is essential to sketch both the solid and its cross section to enhance understanding and avoid errors in calculations.

As you continue practicing these volume problems, remember to evaluate the integral to find the numerical value for the volume of the solid. This foundational approach will be crucial as you tackle more complex three-dimensional shapes in future lessons.

Introduction to Cross Sections Example 1

In this discussion, we explore the evaluation of a volume integral for a three-dimensional shape formed by extending a curve upwards, where the cross sections are squares. The area of a square is calculated as the side length squared, and in this case, the side length is represented by the function \( f(x) \). Therefore, the volume integral is set up as the integral of \( f(x)^2 \).

To evaluate the integral, we first square the function, resulting in the expression:

\[ \int_{0}^{2} (16 - 8x^2 + x^4) \, dx \]

Next, we find the antiderivative of each term within the integral:

\[ 16x - \frac{8}{3}x^3 + \frac{1}{5}x^5 \bigg|_{0}^{2} \]

Applying the Fundamental Theorem of Calculus, we substitute the upper and lower bounds into the antiderivative:

\[ 16(2) - \frac{8}{3}(2^3) + \frac{1}{5}(2^5) - (0) \]

Calculating each term gives:

• \( 16 \times 2 = 32 \)
• \( 2^3 = 8 \) so \( \frac{8}{3} \times 8 = \frac{64}{3} \)
• \( 2^5 = 32 \) so \( \frac{1}{5} \times 32 = \frac{32}{5} \)

Thus, we have:

\[ 32 - \frac{64}{3} + \frac{32}{5} \]

To combine these fractions, we find a common denominator of 15:

• \( 32 = \frac{480}{15} \)
• \( \frac{64}{3} = \frac{320}{15} \)
• \( \frac{32}{5} = \frac{96}{15} \)

Now, substituting these values back, we get:

\[ \frac{480}{15} - \frac{320}{15} + \frac{96}{15} = \frac{256}{15} \]

This fraction represents the volume of the three-dimensional shape with square cross sections. If desired, this can also be expressed as a decimal, yielding approximately \( 17.07 \). This volume is significant as it quantifies the space occupied by the shape defined by the given function.

If there are any questions regarding the setup or evaluation of this integral, further clarification is encouraged.

Find the volume of the solid whose base is the region bounded by the function $f\left(x\right)=\sqrt{9-x^2}$ and the x-axis with square cross sections perpendicular to the x-axis.

• A. 54
• B. 36
• C. 18
• D. 72

Answer: (B)

Introduction to Cross Sections Example 2

To determine the volume of a solid with a base defined by the function \( y = 4 - x^2 \) over the interval from \( x = 0 \) to \( x = 2 \), we first visualize the solid and its cross sections. The base area is bounded by the curve, and the solid extends vertically from this base. When we slice the solid parallel to the y-axis, we obtain cross sections that can take different shapes.

For the first scenario, the cross sections are isosceles right triangles. The area \( A \) of an isosceles right triangle can be expressed as:

\[A = \frac{1}{2} \times \text{base} \times \text{height}\]

In this case, both the base and height of the triangle are equal to the function \( f(x) = 4 - x^2 \). Therefore, the area function becomes:

\[A = \frac{1}{2} \times f(x) \times f(x) = \frac{1}{2} f(x)^2 = \frac{1}{2} (4 - x^2)^2\]

To find the volume \( V \) of the solid, we set up the definite integral of the area function from \( x = 0 \) to \( x = 2 \):

\[V = \int_{0}^{2} A \, dx = \int_{0}^{2} \frac{1}{2} (4 - x^2)^2 \, dx\]

Next, we consider the second scenario where the cross sections are circular. The area \( A \) of a circle is given by:

\[A = \pi r^2\]

Here, the diameter of the circle is equal to the function \( f(x) \), which means the radius \( r \) is half of that:

\[r = \frac{1}{2} f(x) = \frac{1}{2} (4 - x^2)\]

Substituting this into the area formula gives:

\[A = \pi \left( \frac{1}{2} f(x) \right)^2 = \pi \left( \frac{1}{2} (4 - x^2) \right)^2 = \frac{\pi}{4} (4 - x^2)^2\]

To find the volume \( V \) for this case, we again set up the definite integral from \( x = 0 \) to \( x = 2 \):

\[V = \int_{0}^{2} A \, dx = \int_{0}^{2} \frac{\pi}{4} (4 - x^2)^2 \, dx\]

These integrals represent the volumes of the solid with isosceles right triangle and circular cross sections, respectively. By evaluating these integrals, one can find the exact volumes of the solids formed by the given base function.

Find the volume for a solid whose base is the region between the curve $y=\sqrt{\sin x}$ and the x-axis on the interval from $[0,\pi ]$ and whose cross sections are equilateral triangles with bases parallel to the y-axis.

• A. $\frac{\sqrt{3}}{2}$
• B. $-\frac{\sqrt{3}}{4}$
• C. $-\frac{\sqrt{3}}{2}$
• D. $-\frac{4+\sqrt{3}}{4}$

Answer: (A)

Finding Volume Using Disks

To find the volume of a three-dimensional solid formed by rotating a function around an axis, we utilize the concept of solids of revolution. In this context, the cross-sections of the solid are circular, which simplifies the process of determining the area function we need to integrate. The area of a circle is given by the formula \( A = \pi r^2 \), where \( r \) is the radius of the circle.

Consider the function \( f(x) = 4 - x^2 \) defined on the interval from \(-2\) to \(2\). When this function is revolved around the x-axis, it creates a solid of revolution. The radius \( r \) of the circular cross-section at any point \( x \) is determined by the distance from the curve to the axis of rotation. In this case, since we are revolving around the x-axis (where \( y = 0 \)), the radius function is simply \( r(x) = f(x) = 4 - x^2 \).

To set up the volume integral, we integrate the area of the circular cross-section from the lower limit to the upper limit of the interval. The volume \( V \) can be expressed as:

\( V = \int_{-2}^{2} \pi (r(x))^2 \, dx = \int_{-2}^{2} \pi (4 - x^2)^2 \, dx \)

This integral represents the volume of the solid formed by revolving the function around the x-axis. To find the numerical value of the volume, one would evaluate this integral. The process involves expanding the integrand, integrating term by term, and then applying the limits of integration.

Understanding the relationship between the function, its area, and the resulting volume is crucial in mastering the concept of solids of revolution and the disk method. This foundational knowledge will be beneficial as you continue to explore more complex shapes and integration techniques in calculus.

Finding Volume Using Disks Example 3

To find the volume of a solid formed by revolving a function around the x-axis, we utilize the method of disks. The volume can be calculated using the integral of the area of circular cross-sections, where the area \( A \) of a circle is given by the formula \( A = \pi r^2 \). In this case, the radius \( r \) is defined by the function \( r(x) = 4 - x^2 \), leading to the volume integral:

$$ V = \int_{-2}^{2} \pi (4 - x^2)^2 \, dx $$

First, we expand the integrand:

$$ (4 - x^2)^2 = 16 - 8x^2 + x^4 $$

Thus, the volume integral becomes:

$$ V = \pi \int_{-2}^{2} (16 - 8x^2 + x^4) \, dx $$

Next, we can factor out the constant \( \pi \) from the integral:

$$ V = \pi \left( \int_{-2}^{2} 16 \, dx - 8 \int_{-2}^{2} x^2 \, dx + \int_{-2}^{2} x^4 \, dx \right) $$

Calculating each integral separately, we find:

1. For \( \int_{-2}^{2} 16 \, dx \):

$$ = 16 \cdot (2 - (-2)) = 16 \cdot 4 = 64 $$

2. For \( \int_{-2}^{2} x^2 \, dx \):

Using the power rule, we have:

$$ \int x^2 \, dx = \frac{x^3}{3} $$

Evaluating from -2 to 2 gives:

$$ \left[ \frac{(2)^3}{3} - \frac{(-2)^3}{3} \right] = \frac{8}{3} - \left(-\frac{8}{3}\right) = \frac{16}{3} $$

3. For \( \int_{-2}^{2} x^4 \, dx \):

Using the power rule again:

$$ \int x^4 \, dx = \frac{x^5}{5} $$

Evaluating from -2 to 2 gives:

$$ \left[ \frac{(2)^5}{5} - \frac{(-2)^5}{5} \right] = \frac{32}{5} - \left(-\frac{32}{5}\right) = \frac{64}{5} $$

Now substituting these results back into the volume integral:

$$ V = \pi \left( 64 - 8 \cdot \frac{16}{3} + \frac{64}{5} \right) $$

Calculating the terms:

$$ 8 \cdot \frac{16}{3} = \frac{128}{3} $$

Combining the terms requires a common denominator, which is 15:

1. Convert \( 64 \) to fifteenths: \( 64 = \frac{960}{15} \)

2. Convert \( \frac{128}{3} \) to fifteenths: \( \frac{128}{3} = \frac{640}{15} \)

3. Convert \( \frac{64}{5} \) to fifteenths: \( \frac{64}{5} = \frac{192}{15} \)

Now we can combine:

$$ V = \pi \left( \frac{960}{15} - \frac{640}{15} + \frac{192}{15} \right) $$

$$ = \pi \left( \frac{512}{15} \right) $$

Thus, the final volume of the solid of revolution is:

$$ V = \frac{512\pi}{15} $$

This volume represents the space occupied by the solid formed by revolving the given function around the x-axis. If you have any questions about the setup or calculations, feel free to ask for clarification.

Finding Volume Using Disks Example 4

To find the volume of a solid of revolution formed by rotating the area between the function \( f(x) = 4 - x^2 \) and the line \( y = 2 \) around the line \( y = 2 \), we first visualize the solid. The area between the curve and the line is revolved around \( y = 2 \), creating a three-dimensional shape resembling an oval.

To set up the volume integral, we need to determine the radius function \( r(x) \) and the bounds of integration. The radius is defined as the distance from the curve to the axis of rotation. Here, the radius function is given by:

\[ r(x) = f(x) - 2 = (4 - x^2) - 2 = 2 - x^2 \]

Next, we need to find the bounds of the integral, which correspond to the points where the function intersects the line \( y = 2 \). To find these intersection points, we set the function equal to 2:

\[ 4 - x^2 = 2 \]

Subtracting 4 from both sides gives:

\[ -x^2 = -2 \]

Thus, we have:

\[ x^2 = 2 \]

Taking the square root yields the intersection points:

\[ x = \pm \sqrt{2} \]

With the bounds established as \( x = -\sqrt{2} \) and \( x = \sqrt{2} \), we can now set up the volume integral. The volume \( V \) of the solid of revolution is given by:

\[ V = \int_{-\sqrt{2}}^{\sqrt{2}} \pi [r(x)]^2 \, dx = \int_{-\sqrt{2}}^{\sqrt{2}} \pi (2 - x^2)^2 \, dx \]

This integral represents the volume of the solid formed by the revolution. Evaluating this integral will yield the numerical volume of the solid.

Find the volume of the solid obtained by rotating the region bounded by $y=x+4$, $y=0$, $x=1$ & $x=5$.

• A. $\frac{854}{3}\pi$
• B. $4\pi$
• C. $\frac{604}{3}\pi$
• D. $604\pi$

Answer: (C)

Disk Method Using y-Axis

To find the volume of a solid of revolution formed by rotating a function around an axis, we can use the disk method. This method involves integrating the area of circular cross-sections of the solid. When revolving a function around the x-axis, the radius of the circular cross-section is a function of x, and we integrate with respect to x. However, when revolving around the y-axis, the radius becomes a function of y, necessitating a change in our approach.

Consider the function \( f(x) = 4 - x^2 \) defined on the interval from 0 to 2. When this function is revolved around the y-axis, we first need to express it in terms of y. Setting \( y = 4 - x^2 \), we can rearrange this to find \( x \) in terms of \( y \):

\[x^2 = 4 - y \implies x = \sqrt{4 - y}\]

Since we are only interested in the positive square root (as we are in the first quadrant), we define \( g(y) = \sqrt{4 - y} \). This function represents the radius of our circular cross-section as a function of y.

Next, we determine the bounds of our integral. The original function is defined from \( x = 0 \) to \( x = 2 \). To find the corresponding y-values, we evaluate the function at these x-values:

For \( x = 0 \):

\[y = 4 - 0^2 = 4\]

For \( x = 2 \):

\[y = 4 - 2^2 = 0\]

Thus, the bounds for our integral are from \( y = 0 \) to \( y = 4 \).

Now we can set up the volume integral. The volume \( V \) of the solid formed by revolving the function around the y-axis is given by:

\[V = \int_{0}^{4} \pi [r(y)]^2 \, dy\]

Substituting our radius function \( r(y) = \sqrt{4 - y} \), we have:

\[V = \int_{0}^{4} \pi (\sqrt{4 - y})^2 \, dy\]

This simplifies to:

\[V = \int_{0}^{4} \pi (4 - y) \, dy\]

To evaluate this integral, you would proceed with the integration process, which will yield the volume of the solid formed by the revolution of the function around the y-axis. The final result will provide the numerical value representing the volume of the solid.

Disk Method Using y-Axis Example 5

To find the volume of a solid formed by rotating the function \( f(x) = 4 - x^2 \) over the interval from \( x = 0 \) to \( x = 2 \) about the y-axis, we first need to express the function in terms of \( y \). This allows us to determine the radius of the solid as a function of \( y \) and set up the volume integral.

The volume \( V \) of the solid can be calculated using the formula:

\[V = \int_{a}^{b} \pi r^2 \, dy\]

In this case, we rewrite the function to find the bounds for \( y \). The function intersects the y-axis at \( y = 4 \) when \( x = 0 \) and at \( y = 0 \) when \( x = 2 \). Thus, the bounds for our integral are from \( y = 0 \) to \( y = 4 \).

Next, we set up the integral:

\[V = \int_{0}^{4} \pi \sqrt{4 - y^2} \, dy\]

We can simplify this integral by factoring out the constant \( \pi \):

\[V = \pi \int_{0}^{4} (4 - y) \, dy\]

Now, we find the antiderivative of \( 4 - y \):

\[\int (4 - y) \, dy = 4y - \frac{1}{2}y^2\]

We evaluate this from \( 0 \) to \( 4 \):

\[V = \pi \left[ 4(4) - \frac{1}{2}(4^2) \right] - \pi \left[ 4(0) - \frac{1}{2}(0^2) \right]\]

Calculating the upper bound:

\[4(4) = 16 \quad \text{and} \quad \frac{1}{2}(4^2) = 8\]

Thus, we have:

\[V = \pi (16 - 8) = \pi (8)\]

Therefore, the volume of the solid is:

\[V = 8\pi\]

This result represents the volume of the dome-shaped solid formed by the rotation of the function about the y-axis.

Disk Method Using y-Axis Example 6

To find the volume of the solid of revolution formed by rotating the area between the function \( f(x) = 4 - x^2 \), the line \( y = 4 \), and the line \( x = 2 \) around the line \( x = 2 \) in the first quadrant, we start by visualizing the solid and its cross-sections. The area of interest is bounded above by the line \( y = 4 \) and below by the curve \( f(x) \). When this area is revolved around the line \( x = 2 \), it creates a funnel-like shape.

The cross-sections of this solid are circular, and the radius of each circular cross-section is determined by the distance from the curve to the axis of revolution. Since we are revolving around the vertical line \( x = 2 \), we need to express the radius as a function of \( y \). To do this, we first rewrite the function in terms of \( y \): starting from \( y = 4 - x^2 \), we rearrange it to find \( x \) as follows:

\( y = 4 - x^2 \) implies \( x^2 = 4 - y \), leading to \( x = \sqrt{4 - y} \) (considering only the positive root since we are in the first quadrant).

The radius \( r \) of the circular cross-section is then given by the distance from the curve to the line \( x = 2 \):

\( r(y) = 2 - \sqrt{4 - y} \).

Next, we determine the bounds for our integral. Since we are working in the first quadrant and bounded by \( y = 4 \), the upper bound is \( y = 4 \) and the lower bound is \( y = 0 \). The volume \( V \) of the solid can be calculated using the formula for the area of a circle, \( A = \pi r^2 \), integrated with respect to \( y \):

\( V = \int_{0}^{4} \pi (r(y))^2 \, dy = \int_{0}^{4} \pi (2 - \sqrt{4 - y})^2 \, dy. \)

Expanding the integrand, we have:

\( (2 - \sqrt{4 - y})^2 = 4 - 4\sqrt{4 - y} + (4 - y) = 8 - y - 4\sqrt{4 - y}. \)

Thus, the volume integral becomes:

\( V = \pi \int_{0}^{4} (8 - y - 4\sqrt{4 - y}) \, dy. \)

This integral can be split into three separate integrals:

\( V = \pi \left( \int_{0}^{4} 8 \, dy - \int_{0}^{4} y \, dy - 4 \int_{0}^{4} \sqrt{4 - y} \, dy \right). \)

Calculating each integral separately:

1. For \( \int_{0}^{4} 8 \, dy = 8y \big|_{0}^{4} = 32 \).

2. For \( \int_{0}^{4} y \, dy = \frac{y^2}{2} \big|_{0}^{4} = \frac{16}{2} = 8 \).

3. For \( \int_{0}^{4} \sqrt{4 - y} \, dy \), we use the substitution \( u = 4 - y \), which gives \( du = -dy \). Changing the limits accordingly, when \( y = 0 \), \( u = 4 \) and when \( y = 4 \), \( u = 0 \). Thus, the integral becomes:

\( -\int_{4}^{0} u^{1/2} \, du = \int_{0}^{4} u^{1/2} \, du = \frac{2}{3} u^{3/2} \big|_{0}^{4} = \frac{2}{3} (8) = \frac{16}{3}. \)

Putting it all together, we have:

\( V = \pi \left( 32 - 8 - 4 \cdot \frac{16}{3} \right) = \pi \left( 32 - 8 - \frac{64}{3} \right) = \pi \left( \frac{96}{3} - \frac{64}{3} \right) = \pi \left( \frac{32}{3} \right). \)

Thus, the final volume of the solid of revolution is:

\( V = \frac{32\pi}{3}. \)

Find the volume of the solid formed by revolving the area bounded by $f\left(x\right)=x^2$ from $x=0$ to $x=3$ and the y-axis around the y-axis.

• A. $81\pi$
• B. $162\pi$
• C. $\frac{81}{2}\pi$
• D. $324\pi$

Answer: (C)

Here’s what students ask on this topic:

What is the disk method in calculus for finding volume?

The disk method is a technique used in calculus to find the volume of a solid of revolution. This method involves revolving a function around an axis, typically the x-axis or y-axis, to create a three-dimensional shape. The cross-sections of this solid are circular, resembling disks. To calculate the volume, you integrate the area of these circular cross-sections, which is given by the formula $\pi r^2$, where $r$ is the radius of the disk. The radius is determined by the distance from the curve to the axis of revolution. The integral is set up over the interval of the function, and the result gives the total volume of the solid.

How do you find the volume of a solid with square cross sections?

To find the volume of a solid with square cross sections, you first need to determine the area of one square cross section. If the base of the solid is defined by a function $f\left(x\right)$, the side length of the square is $f\left(x\right)$. Therefore, the area of the square is $f^{\left(}$. To find the volume, integrate this area function over the interval of the base. The integral is set up as ${\int }_a^b{f}^{\left(}dx$, where $a$ and $b$ are the bounds of the interval. The result of this integral gives the total volume of the solid.

What is a solid of revolution in calculus?

A solid of revolution is a three-dimensional shape created by rotating a two-dimensional curve around a specific axis. This process generates a symmetrical solid, where the cross-sections perpendicular to the axis of rotation are circular. The volume of such a solid can be calculated using methods like the disk or washer method, depending on the specific setup of the problem. The disk method involves integrating the area of circular cross-sections, while the washer method is used when there is a hollow region within the solid. These methods are essential in calculus for solving problems related to volumes of solids formed by rotation.

How do you set up an integral for the volume of a solid of revolution?

To set up an integral for the volume of a solid of revolution, first identify the axis of rotation and the function being revolved. If revolving around the x-axis, the radius of the circular cross-section is the distance from the curve to the axis, given by the function $f\left(x\right)$. The area of the cross-section is $\pi f^{\left(}$. Integrate this area over the interval of the function: ${\int }_a^b\pi f^{\left(}dx$. If revolving around the y-axis, express the function in terms of $y$ and integrate with respect to $y$. The integral provides the volume of the solid.

What is the difference between the disk method and the washer method?

The disk method and the washer method are both techniques used to find the volume of solids of revolution, but they differ in their application. The disk method is used when the solid has no hollow region, meaning the cross-sections are solid disks. The volume is calculated by integrating the area of these disks, $\pi r^2$, where $r$ is the radius. The washer method, on the other hand, is used when there is a hollow region within the solid, resulting in washer-shaped cross-sections. The volume is found by integrating the difference between the outer and inner radii, $\pi (R^2-r^2)$, where $R$ is the outer radius and $r$ is the inner radius.