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Using definite integrals to find volume

Just as we can use definite integrals to add the areas of rectangular slices to find the exact area that lies between two curves, we can also use integrals to find the volume of regions whose cross-sections have a…

Introduction

In particular, we can determine the volume of solids whose cross-sections are all thin cylinders (or washers) by adding up the volumes of these individual slices. We first consider a familiar shape in the following Preview Activity: a circular cone.

Exploration

The Volume of a Solid of Revolution

A solid of revolution is a three-dimensional solid that can be generated by revolving one or more curves around a fixed axis. For example, the circular cone in Preview Activity is the solid of revolution generated by revolving the portion of the line \(y = 3 - \frac{3}{5}x\) from \(x = 0\) to \(x = 5\) about the \(x\)-axis. Notice that if we slice a solid of revolution perpendicular to the axis of revolution, the resulting cross-section is a circle.

We first consider solids whose slices are thin cylinders. Recall that the volume of a cylinder is given by \(V = \pi r^2 h\).

Example

Find the volume of the solid of revolution generated when the region \(R\) bounded by \(y = 4-x^2\) and the \(x\)-axis is revolved about the \(x\)-axis.

Solution

First, we observe that \(y = 4-x^2\) intersects the \(x\)-axis at the points \((-2,0)\) and \((2,0)\). When we revolve the region \(R\) about the \(x\)-axis, we get the three-dimensional solid pictured in Figure.

We slice the solid into vertical slices of thickness \(\Delta x\) between \(x = -2\) and \(x = 2\). A representative slice is a cylinder of height \(\Delta x\) and radius \(4-x^2\). Hence, the volume of the slice is \[\begin{aligned}\end{aligned}\].

Using a definite integral to sum the volumes of the representative slices, it follows that \[\begin{aligned}\end{aligned}\].

It is straightforward to evaluate the integral and find that the volume is \(V = \frac{512}{15}\pi\).

For a solid such as the one in Example, where each slice is a cylindrical disk, we first find the volume of a typical slice (noting particularly how this volume depends on \(x\)), and then integrate over the range of \(x\)-values that bound the solid. Often, we will be content with simply finding the integral that represents the volume; if we desire a numeric value for the integral, we typically use a calculator or computer algebra system to find that value.

This method for finding the volume of a solid of revolution is often called the disk method.

If \(y = r(x)\) is a nonnegative continuous function on \([a,b]\), then the volume of the solid of revolution generated by revolving the curve about the \(x\)-axis over this interval is given by \[\begin{aligned}\end{aligned}\].

A different type of solid can emerge when two curves are involved, as we see in the following example.

Condensed — the full section is in Boelkins, Active Calculus.

Revolving about the y-axis

When we revolve a given region about the \(y\)-axis, the representative slices now have thickness \(\Delta y\), which means that we must integrate with respect to \(y\).

Example

Find the volume of the solid of revolution generated when the finite region \(R\) that lies between \(y = \sqrt{x}\) and \(y = x^4\) is revolved about the \(y\)-axis.

Solution

These two curves intersect at \(x = 0\) and \(x = 1\), hence at the points \((0,0)\) and \((1,1)\). When we revolve the region \(R\) about the \(y\)-axis, we get the three-dimensional solid pictured at left in Figure.

Note that the slices are cylindrical washers only if taken perpendicular to the \(y\)-axis. We slice the solid horizontally, starting at \(y = 0\) and proceeding up to \(y = 1\). The thickness of a representative slice is \(\Delta y\), so we must express the integrand in terms of \(y\). The inner radius is determined by the curve \(y = \sqrt{x}\), so we solve for \(x\) and get \(x = y^2 = r(y)\). In the same way, we solve the curve \(y = x^4\) (which governs the outer radius) for \(x\) in terms of \(y\), and hence \(x = \sqrt[4]{y}\). Therefore, the volume of a typical slice is \[\begin{aligned}\end{aligned}\].

We use a definite integral to sum the volumes of all the slices from \(y = 0\) to \(y = 1\). The total volume is \[\begin{aligned}\end{aligned}\].

It is straightforward to evaluate the integral and find that \(V = \frac{7}{15} \pi\).

Revolving about horizontal and vertical lines other than the coordinate axes

It is possible to revolve a region around any horizontal or vertical line. Doing so adjusts the radii of the cylinders or washers involved by a constant value. A careful, well-labeled plot of the solid of revolution will usually reveal how the different axis of revolution affects the definite integral.

Example

Find the volume of the solid of revolution generated when the finite region \(S\) that lies between \(y = x^2\) and \(y = x\) is revolved about the line \(y = -1\).

Solution

Graphing the region between the two curves in the first quadrant between their points of intersection (\((0,0)\) and \((1,1)\)) and then revolving the region about the line \(y = -1\), we see the solid shown in Figure. Each slice of the solid perpendicular to the axis of revolution is a washer, and the radii of each washer are governed by the curves \(y = x^2\) and \(y = x\). But we also see that there is one added change: the axis of revolution adds a fixed length to each radius. The inner radius of a typical slice, \(r(x)\), is given by \(r(x) = x^2 + 1\), while the outer radius is \(R(x) = x+1\).

Therefore, the volume of a typical slice is \[\begin{aligned}\end{aligned}\].

Finally, we integrate to find the total volume, and \[\begin{aligned}\end{aligned}\].

Summary

We can use a definite integral to find the volume of a three-dimensional solid of revolution that results from revolving a two-dimensional region about a particular axis by taking slices perpendicular to the axis of revolution which will then be circular disks or washers.
If we revolve about a vertical line and slice perpendicular to that line, then our slices are horizontal and of thickness \(\Delta y\). This leads us to integrate with respect to \(y\), as opposed to with respect to \(x\) when we slice a solid vertically.
If we revolve about a line other than the \(x\)- or \(y\)-axis, we need to carefully account for the shift that occurs in the radius of a typical slice. Normally, this shift involves taking a sum or difference of the function along with the constant connected to the equation for the horizontal or vertical line; a well-labeled diagram is usually the best way to decide the new expression for the radius.

Practice (6)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Consider the curve \(f(x) = 3 \cos(\frac{x^3}{4})\) and the portion of its graph that lies in the first quadrant between the \(y\)-axis and the first positive value of \(x\) for which \(f(x) = 0\). Let \(R\) denote the region bounded by this portion of \(f\), the \(x\)-axis, and the \(y\)-axis.
1. Set up a definite integral whose value is the exact arc length of \(f\) that lies along the upper boundary of \(R\). Use technology appropriately to evaluate the integral you find.
2. Set up a definite integral whose value is the exact area of \(R\). Use technology appropriately to evaluate the integral you find.
3. Suppose that the region \(R\) is revolved around the \(x\)-axis. Set up a definite integral whose value is the exact volume of the solid of revolution that is generated. Use technology appropriately to evaluate the integral you find.
4. Suppose instead that \(R\) is revolved around the \(y\)-axis. If possible, set up an integral expression whose value is the exact volume of the solid of revolution and evaluate the integral using appropriate technology. If not possible, explain why.
เปิดเผยคำตอบ
1. Plotting the function and using technology to find where \(f(x)\) intersects the \(x\)-axis, we see that the region \(R\) as pictured in the following figure.
  To apply the formula we developed for arc length, we need the derivative of \(f\). By the chain rule, \(f'(x) = -3 \sin(\frac{x^3}{4}) \cdot \frac{3}{4}x^2\), so the arc length of \(f\) that lies along the upper boundary of \(R\) is \[\begin{aligned}\end{aligned}\]. Using technology to evaluate the integral, we find \(L \approx 4.10521\).
2. The area of \(R\) is \[\begin{aligned}\end{aligned}\].
3. Revolving \(R\) around the \(x\)-axis, we can use the disk method with slices perpendicular to the \(x\)-axis to find the volume of the resulting region. Noting that the radius of each such disk is \(r(x) = 3 \cos(\frac{x^3}{4})\) and the width of each such disk is \(\triangle x\), we know the volume of each slice is \[\begin{aligned}\end{aligned}\] and thus the total volume is \[\begin{aligned}\end{aligned}\]
4. If we instead revolve \(R\) around the \(y\)-axis, we need to use horizontal disks of thickness \(\triangle y\) in order to apply the disk method. To do that, we also need to express \(x\) as a function of \(y\), rather than \(y\) as a function of \(x\) as given. Thus we take the given function \(y = 3 \cos(\frac{x^3}{4})\) and solve the equation for \(x\). First, we observe \[\begin{aligned}\end{aligned}\] and thus using the inverse cosine, \[\begin{aligned}\end{aligned}\] and therefore \(x^3 = 4\arccos \left(\frac{y}{3} \right)\) so that \[\begin{aligned}\end{aligned}\]. Now, a typical slice has radius \(r(y) = \sqrt[3]{4\arccos \left(\frac{y}{3} \right)}\) with thickness \(\triangle y\), and volume \[\begin{aligned}\end{aligned}\]. Integrating to add up the slices from \(y = 0\) to \(y = 3\), we have the volume of the solid of revolution is \[\begin{aligned}\end{aligned}\].
Consider the curves given by \(y = \sin(x)\) and \(y = \cos(x)\). For each of the following problems, you should include a sketch of the region/solid being considered, as well as a labeled representative slice.
1. Sketch the region \(R\) bounded by the \(y\)-axis and the curves \(y = \sin(x)\) and \(y = \cos(x)\) up to the first positive value of \(x\) at which they intersect. What is the exact intersection point of the curves?
2. Set up a definite integral whose value is the exact area of \(R\).
3. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving \(R\) about the \(x\)-axis.
4. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving \(R\) about the \(y\)-axis.
5. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving \(R\) about the line \(y = 2\).
6. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving \(R\) about the line \(x = -1\).
เปิดเผยคำตอบ
1. See the sketch above. The exact intersection point of the curves is where \(\cos(x) = \sin(x)\) in the first quadrant, and thus at the point \((\frac{\pi}{4}, \frac{\sqrt{2}}{2})\).
2. The area of \(R\) is given by the definite integral \[\begin{aligned}\end{aligned}\].
3. Revolving \(R\) about the \(x\)-axis and using slices perpendicular to the axis with thickness \(\triangle x\), the outer radius of the resulting washers is \(R(x) = \cos(x)\), while the inner radius of the washers is \(r(x) = \sin(x)\). The volume of a typical slice is \[\begin{aligned}\end{aligned}\] and thus using an integral to sum the slices over the interval \(x = 0\) to \(x = \pi/4\), we find the volume of the solid is \[\begin{aligned}\end{aligned}\].
4. Revolving \(R\) about the \(y\)-axis, the resulting figure has solid cross sections that are disks perpendicular to the \(y\)-axis with thickness \(\triangle x\), but the radius \(R(y)\) that governs the radius of each disk changes at the height where \(y = \frac{\sqrt{2}}{2}\) in the interval from \(y = 0\) to \(y = 1\). We also need to express the two curves that bound the region with \(x\) as a function of \(y\). On the interval from \(y = 0\) to \(y = \frac{\sqrt{2}}{2}\), the sine function provides the boundary, which we write as \(x = \arcsin(y)\); on the interval from \(y = \frac{\sqrt{2}}{2}\) to \(y = 1\), the cosine function provides the boundary, which we write as \(x = \arccos(y)\). On the lower interval, a typical slice has volume \[\begin{aligned}\end{aligned}\]. Using similar reasoning for the upper interval and summing the slices using appropriate definite integrals, we find that \[\begin{aligned}\end{aligned}\]
5. Draw a picture. Revolving \(R\) about the line \(y = 2\), we see that cross-sections perpendicular to the line \(y = 2\) are washers with thickness \(\triangle x\), outer radius \(R(x) = 2 - \sin(x)\), and inner radius \(r(x) = 2 - \cos(x)\), since \(2 - \sin(x) \ge 2 - \cos(x)\) on the interval \([0, \frac{\pi}{4}]\). Thus, the volume of a typical slice is \[\begin{aligned}\end{aligned}\] and hence integrating to sum the total volume, we find \[\begin{aligned}\end{aligned}\].
6. Revolving \(R\) about the line \(x = -1\), the resulting figure has solid cross sections that are washers perpendicular to the \(y\)-axis with thickness \(\triangle x\); the inner radius is always \(r(y) = 1\), but the outer radius \(R(y)\) that governs the larger radius of each washer changes at the height where \(y = \frac{\sqrt{2}}{2}\) in the interval from \(y = 0\) to \(y = 1\). As in (d), we express the two curves that bound the region with \(x\) as a function of \(y\). On the interval from \(y = 0\) to \(y = \frac{\sqrt{2}}{2}\), \(R(y) = 1 + \arcsin(y)\); on the interval from \(y = \frac{\sqrt{2}}{2}\) to \(y = 1\),\(R(y) = 1 + \arccos(y)\). On the lower interval, a typical slice has volume \[\begin{aligned}\end{aligned}\]. Using similar reasoning for the upper interval and summing the slices using appropriate definite integrals, we find that \[\begin{aligned}\end{aligned}\]
Consider the finite region \(R\) that is bounded by the curves \(y = 1+\frac{1}{2}(x-2)^2\), \(y=\frac{1}{2}x^2\), and \(x = 0\).
1. Determine a definite integral whose value is the area of the region enclosed by the two curves.
2. Find an expression involving one or more definite integrals whose value is the volume of the solid of revolution generated by revolving the region \(R\) about the line \(y = -1\).
3. Determine an expression involving one or more definite integrals whose value is the volume of the solid of revolution generated by revolving the region \(R\) about the \(y\)-axis.
4. Find an expression involving one or more definite integrals whose value is the perimeter of the region \(R\).
เปิดเผยคำตอบ
1. A picture of the region is shown in the figure above. The point of intersection occurs when \[\begin{aligned}\end{aligned}\], so \(2+(x-2)^2 = x^2\), which implies \(2+(x^2-4x+4) = x^2\) and thus \(4x=6\) so \(x = \frac{3}{2}\). Since \(g(x) = 1+\frac{1}{2}(x-2)^2 \ge \frac{1}{2}x^2 = f(x)\) on the interval \(\left[0, \frac{3}{2} \right]\), the area of the region between the two curves is \[\begin{aligned}A = &= \int_0^{1.5} 1+\frac{1}{2}(x-2)^2 - \frac{1}{2}x^2 \ dx \\ &= \int_0^{1.5} 1+\frac{1}{2}(x^2-4x+4) - \frac{1}{2}x^2 \ dx \\ &= \int_0^{1.5} 3-2x \ dx \\ &= \left(3x-x^2\right)\biggm|_{0}^{1.5} \\ &= 4.5-2.25 \\ &= 2.25\end{aligned}\].
2. (You should draw a picture of the solid, based on the figure above, with the region \(R\) revolved about \(y = -1\).) We slice vertically so that our cross sections are washers with thickness \(\triangle x\). Because we are revolving about \(y = -1\), the outer radius is \(R(x) = 2+\frac{1}{2}(x-2)^2\), while the inner radius is \(1+\frac{1}{2}x^2\). Hence the volume of a typical slice, which has thickness \(\triangle x\) is given by \[\begin{aligned}\end{aligned}\]. Hence, the volume of the solid obtained by revolving the region \(R\) around the line \(y=-1\) is given by \[\begin{aligned}\end{aligned}\]
3. (Draw a picture of the solid, based on revolving the region \(R\) about the \(y\)-axis.) We slice horizontally and thus the resulting cross sections are cylinders with thickness \(\triangle y\). The radius of the disks changes as \(y\) passes through \(f\left(\frac{3}{2}\right) = \frac{9}{8} = 1.125\), so we need two different integrals to find the total volume. Since we are slicing horizontally, we will integrate with respect to \(y\), so we need to solve both of our equations for \(x\) as functions of \(y\). Observe that \(y = 1+\frac{1}{2}(x-2)^2\) implies \(2(y-1) = (x-2)^2\), so \(x-2 = \pm \sqrt{2(y-1)}\) and thus \(x = 2 \pm \sqrt{2(y-1)}\), and similarly \(y = \frac{1}{2}x^2\) implies \(x = \pm \sqrt{2y}\). The portion of the curve defined by \(y = 1+\frac{1}{2}(x-2)^2\) that determines one of the boundaries is the left hand portion of the parabola, so \(x = 2 - \sqrt{2(y-1)}\), while the portion of the curve \(y = \frac{1}{2}x^2\) that determines the other boundary is the right hand portion of the parabola, so \(x = \sqrt{2y}\). These two curves define the radius of the disk (one for \(y \lt 1.125\), the other for \(y \gt 1.125\)) and the lowest point on the region is \((0,0)\) while the highest is \((0,3)\). So a sum of integrals that represents the volume of the solid obtained by revolving the region \(R\) around the \(y\)-axis is \[\begin{aligned}\end{aligned}\]
4. Recall that the length of a curve defined by a function \(f\) on an interval \([a,b]\) is given by \[\begin{aligned}\end{aligned}\]. To find the perimeter of the region \(R\) we calculate the length of the curves defined by \(y = 1+\frac{1}{2}(x-2)^2\) (for which \(y' = x-2\)) and \(y=\frac{1}{2}x^2\) (for which \(y' = x\)) on the interval \([0,1.5]\) and the length of the segment along \(x=0\) as \(y\) goes from \(0\) to \(3\). This last length is \(3\), so the perimeter of the region \(R\) is given by \[\begin{aligned}\end{aligned}\].
Consider the curve \(f(x) = 3 \cos(\frac{x^3}{4})\) and the portion of its graph that lies in the first quadrant between the \(y\)-axis and the first positive value of \(x\) for which \(f(x) = 0\). Let \(R\) denote the region bounded by this portion of \(f\), the \(x\)-axis, and the \(y\)-axis.
1. Set up a definite integral whose value is the exact arc length of \(f\) that lies along the upper boundary of \(R\). Use technology appropriately to evaluate the integral you find.
2. Set up a definite integral whose value is the exact area of \(R\). Use technology appropriately to evaluate the integral you find.
3. Suppose that the region \(R\) is revolved around the \(x\)-axis. Set up a definite integral whose value is the exact volume of the solid of revolution that is generated. Use technology appropriately to evaluate the integral you find.
4. Suppose instead that \(R\) is revolved around the \(y\)-axis. If possible, set up an integral expression whose value is the exact volume of the solid of revolution and evaluate the integral using appropriate technology. If not possible, explain why.
เปิดเผยคำตอบ
1. Plotting the function and using technology to find where \(f(x)\) intersects the \(x\)-axis, we see that the region \(R\) as pictured in the following figure.
  To apply the formula we developed for arc length, we need the derivative of \(f\). By the chain rule, \(f'(x) = -3 \sin(\frac{x^3}{4}) \cdot \frac{3}{4}x^2\), so the arc length of \(f\) that lies along the upper boundary of \(R\) is \[\begin{aligned}\end{aligned}\]. Using technology to evaluate the integral, we find \(L \approx 4.10521\).
2. The area of \(R\) is \[\begin{aligned}\end{aligned}\].
3. Revolving \(R\) around the \(x\)-axis, we can use the disk method with slices perpendicular to the \(x\)-axis to find the volume of the resulting region. Noting that the radius of each such disk is \(r(x) = 3 \cos(\frac{x^3}{4})\) and the width of each such disk is \(\triangle x\), we know the volume of each slice is \[\begin{aligned}\end{aligned}\] and thus the total volume is \[\begin{aligned}\end{aligned}\]
4. If we instead revolve \(R\) around the \(y\)-axis, we need to use horizontal disks of thickness \(\triangle y\) in order to apply the disk method. To do that, we also need to express \(x\) as a function of \(y\), rather than \(y\) as a function of \(x\) as given. Thus we take the given function \(y = 3 \cos(\frac{x^3}{4})\) and solve the equation for \(x\). First, we observe \[\begin{aligned}\end{aligned}\] and thus using the inverse cosine, \[\begin{aligned}\end{aligned}\] and therefore \(x^3 = 4\arccos \left(\frac{y}{3} \right)\) so that \[\begin{aligned}\end{aligned}\]. Now, a typical slice has radius \(r(y) = \sqrt[3]{4\arccos \left(\frac{y}{3} \right)}\) with thickness \(\triangle y\), and volume \[\begin{aligned}\end{aligned}\]. Integrating to add up the slices from \(y = 0\) to \(y = 3\), we have the volume of the solid of revolution is \[\begin{aligned}\end{aligned}\].
Consider the curves given by \(y = \sin(x)\) and \(y = \cos(x)\). For each of the following problems, you should include a sketch of the region/solid being considered, as well as a labeled representative slice.
1. Sketch the region \(R\) bounded by the \(y\)-axis and the curves \(y = \sin(x)\) and \(y = \cos(x)\) up to the first positive value of \(x\) at which they intersect. What is the exact intersection point of the curves?
2. Set up a definite integral whose value is the exact area of \(R\).
3. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving \(R\) about the \(x\)-axis.
4. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving \(R\) about the \(y\)-axis.
5. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving \(R\) about the line \(y = 2\).
6. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving \(R\) about the line \(x = -1\).
เปิดเผยคำตอบ
1. See the sketch above. The exact intersection point of the curves is where \(\cos(x) = \sin(x)\) in the first quadrant, and thus at the point \((\frac{\pi}{4}, \frac{\sqrt{2}}{2})\).
2. The area of \(R\) is given by the definite integral \[\begin{aligned}\end{aligned}\].
3. Revolving \(R\) about the \(x\)-axis and using slices perpendicular to the axis with thickness \(\triangle x\), the outer radius of the resulting washers is \(R(x) = \cos(x)\), while the inner radius of the washers is \(r(x) = \sin(x)\). The volume of a typical slice is \[\begin{aligned}\end{aligned}\] and thus using an integral to sum the slices over the interval \(x = 0\) to \(x = \pi/4\), we find the volume of the solid is \[\begin{aligned}\end{aligned}\].
4. Revolving \(R\) about the \(y\)-axis, the resulting figure has solid cross sections that are disks perpendicular to the \(y\)-axis with thickness \(\triangle x\), but the radius \(R(y)\) that governs the radius of each disk changes at the height where \(y = \frac{\sqrt{2}}{2}\) in the interval from \(y = 0\) to \(y = 1\). We also need to express the two curves that bound the region with \(x\) as a function of \(y\). On the interval from \(y = 0\) to \(y = \frac{\sqrt{2}}{2}\), the sine function provides the boundary, which we write as \(x = \arcsin(y)\); on the interval from \(y = \frac{\sqrt{2}}{2}\) to \(y = 1\), the cosine function provides the boundary, which we write as \(x = \arccos(y)\). On the lower interval, a typical slice has volume \[\begin{aligned}\end{aligned}\]. Using similar reasoning for the upper interval and summing the slices using appropriate definite integrals, we find that \[\begin{aligned}\end{aligned}\]
5. Draw a picture. Revolving \(R\) about the line \(y = 2\), we see that cross-sections perpendicular to the line \(y = 2\) are washers with thickness \(\triangle x\), outer radius \(R(x) = 2 - \sin(x)\), and inner radius \(r(x) = 2 - \cos(x)\), since \(2 - \sin(x) \ge 2 - \cos(x)\) on the interval \([0, \frac{\pi}{4}]\). Thus, the volume of a typical slice is \[\begin{aligned}\end{aligned}\] and hence integrating to sum the total volume, we find \[\begin{aligned}\end{aligned}\].
6. Revolving \(R\) about the line \(x = -1\), the resulting figure has solid cross sections that are washers perpendicular to the \(y\)-axis with thickness \(\triangle x\); the inner radius is always \(r(y) = 1\), but the outer radius \(R(y)\) that governs the larger radius of each washer changes at the height where \(y = \frac{\sqrt{2}}{2}\) in the interval from \(y = 0\) to \(y = 1\). As in (d), we express the two curves that bound the region with \(x\) as a function of \(y\). On the interval from \(y = 0\) to \(y = \frac{\sqrt{2}}{2}\), \(R(y) = 1 + \arcsin(y)\); on the interval from \(y = \frac{\sqrt{2}}{2}\) to \(y = 1\),\(R(y) = 1 + \arccos(y)\). On the lower interval, a typical slice has volume \[\begin{aligned}\end{aligned}\]. Using similar reasoning for the upper interval and summing the slices using appropriate definite integrals, we find that \[\begin{aligned}\end{aligned}\]
Consider the finite region \(R\) that is bounded by the curves \(y = 1+\frac{1}{2}(x-2)^2\), \(y=\frac{1}{2}x^2\), and \(x = 0\).
1. Determine a definite integral whose value is the area of the region enclosed by the two curves.
2. Find an expression involving one or more definite integrals whose value is the volume of the solid of revolution generated by revolving the region \(R\) about the line \(y = -1\).
3. Determine an expression involving one or more definite integrals whose value is the volume of the solid of revolution generated by revolving the region \(R\) about the \(y\)-axis.
4. Find an expression involving one or more definite integrals whose value is the perimeter of the region \(R\).
เปิดเผยคำตอบ
1. A picture of the region is shown in the figure above. The point of intersection occurs when \[\begin{aligned}\end{aligned}\], so \(2+(x-2)^2 = x^2\), which implies \(2+(x^2-4x+4) = x^2\) and thus \(4x=6\) so \(x = \frac{3}{2}\). Since \(g(x) = 1+\frac{1}{2}(x-2)^2 \ge \frac{1}{2}x^2 = f(x)\) on the interval \(\left[0, \frac{3}{2} \right]\), the area of the region between the two curves is \[\begin{aligned}A = &= \int_0^{1.5} 1+\frac{1}{2}(x-2)^2 - \frac{1}{2}x^2 \ dx \\ &= \int_0^{1.5} 1+\frac{1}{2}(x^2-4x+4) - \frac{1}{2}x^2 \ dx \\ &= \int_0^{1.5} 3-2x \ dx \\ &= \left(3x-x^2\right)\biggm|_{0}^{1.5} \\ &= 4.5-2.25 \\ &= 2.25\end{aligned}\].
2. (You should draw a picture of the solid, based on the figure above, with the region \(R\) revolved about \(y = -1\).) We slice vertically so that our cross sections are washers with thickness \(\triangle x\). Because we are revolving about \(y = -1\), the outer radius is \(R(x) = 2+\frac{1}{2}(x-2)^2\), while the inner radius is \(1+\frac{1}{2}x^2\). Hence the volume of a typical slice, which has thickness \(\triangle x\) is given by \[\begin{aligned}\end{aligned}\]. Hence, the volume of the solid obtained by revolving the region \(R\) around the line \(y=-1\) is given by \[\begin{aligned}\end{aligned}\]
3. (Draw a picture of the solid, based on revolving the region \(R\) about the \(y\)-axis.) We slice horizontally and thus the resulting cross sections are cylinders with thickness \(\triangle y\). The radius of the disks changes as \(y\) passes through \(f\left(\frac{3}{2}\right) = \frac{9}{8} = 1.125\), so we need two different integrals to find the total volume. Since we are slicing horizontally, we will integrate with respect to \(y\), so we need to solve both of our equations for \(x\) as functions of \(y\). Observe that \(y = 1+\frac{1}{2}(x-2)^2\) implies \(2(y-1) = (x-2)^2\), so \(x-2 = \pm \sqrt{2(y-1)}\) and thus \(x = 2 \pm \sqrt{2(y-1)}\), and similarly \(y = \frac{1}{2}x^2\) implies \(x = \pm \sqrt{2y}\). The portion of the curve defined by \(y = 1+\frac{1}{2}(x-2)^2\) that determines one of the boundaries is the left hand portion of the parabola, so \(x = 2 - \sqrt{2(y-1)}\), while the portion of the curve \(y = \frac{1}{2}x^2\) that determines the other boundary is the right hand portion of the parabola, so \(x = \sqrt{2y}\). These two curves define the radius of the disk (one for \(y \lt 1.125\), the other for \(y \gt 1.125\)) and the lowest point on the region is \((0,0)\) while the highest is \((0,3)\). So a sum of integrals that represents the volume of the solid obtained by revolving the region \(R\) around the \(y\)-axis is \[\begin{aligned}\end{aligned}\]
4. Recall that the length of a curve defined by a function \(f\) on an interval \([a,b]\) is given by \[\begin{aligned}\end{aligned}\]. To find the perimeter of the region \(R\) we calculate the length of the curves defined by \(y = 1+\frac{1}{2}(x-2)^2\) (for which \(y' = x-2\)) and \(y=\frac{1}{2}x^2\) (for which \(y' = x\)) on the interval \([0,1.5]\) and the length of the segment along \(x=0\) as \(y\) goes from \(0\) to \(3\). This last length is \(3\), so the perimeter of the region \(R\) is given by \[\begin{aligned}\end{aligned}\].

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Using definite integrals to find volume

How can we use a definite integral to find the volume of a three-dimensional solid of revolution that results from revolving a two-dimensional region about a particular axis?
In what circumstances do we integrate with respect to y instead of integrating with respect to x?
What adjustments do we need to make if we revolve about a line other than the x- or y-axis?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

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Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

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Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests