Calculus

Applications of Integrals

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Volume of a Solid of Revolution: Disks and Washers

If a region in the plane is revolved about a line in the same plane, the resulting object is known as a solid of revolution.

For example, a solid right circular cylinder can be generated by revolving a rectangle. Similarly, a solid spherical ball can be generated by revolving a semi-disk.

The line about which we rotate the shape is called the axis of revolution.

The Disk Method

The disk method is used when we rotate a single curve \(y = f\left( x \right)\) around the \(x-\) (or \(y-\)) axis.

Suppose that \(y = f\left( x \right)\) is a continuous non-negative function on the interval \(\left[ {a,b} \right].\)

Solid formed by revolving a shape about the x-axis
Figure 1.

The volume of the solid formed by revolving the region bounded by the curve \(y = f\left( x \right)\) and the \(x-\)axis between \(x= a\) and \(x = b\) about the \(x-\)axis is given by

\[V = \pi \int\limits_a^b {{{\left[ {f\left( x \right)} \right]}^2}dx} .\]

The cross section perpendicular to the axis of revolution has the form of a disk of radius \(R = f\left( x \right).\)

Similarly, we can find the volume of the solid when the region is bounded by the curve \(x = f\left( y \right)\) and the \(y-\)axis between \(y = c\) and \(y = d,\) and is rotated about the \(y-\)axis.

Solid formed by revolving a shape about the y-axis
Figure 2.

The resulting formula is

\[V = \pi \int\limits_c^d {{{\left[ {f\left( y \right)} \right]}^2}dy} .\]

The Washer Method

We can extend the disk method to find the volume of a hollow solid of revolution.

Assuming that the functions \(f\left( x \right)\) and \(g\left( x \right)\) are continuous and non-negative on the interval \(\left[ {a,b} \right]\) and \(g\left( x \right) \le f\left( x \right),\) consider a region that is bounded by two curves \(y = f\left( x \right)\) and \(y = g\left( x \right),\) between \(x = a\) and \(x = b.\)

The washer method for finding the volume of a solid generated by revolving about the x-axis.
Figure 3.

The volume of the solid formed by revolving the region about the \(x-\)axis is

\[V = \pi \int\limits_a^b {\left( {{{\left[ {f\left( x \right)} \right]}^2} - {{\left[ {g\left( x \right)} \right]}^2}} \right)dx} .\]

At a point \(x\) on the \(x-\)axis, a perpendicular cross section of the solid is washer-shape with the inner radius \(r = g\left( x \right)\) and the outer radius \(R = f\left( x \right).\)

The volume of the solid generated by revolving about the \(y-\)axis a region between the curves \(x = f\left( y \right)\) and \(x = g\left( y \right),\) where \(g\left( y \right) \le f\left( y \right)\) and \(c \le y \le d\) is given by the formula

\[V = \pi \int\limits_c^d {\left( {{{\left[ {f\left( y \right)} \right]}^2} - {{\left[ {g\left( y \right)} \right]}^2}} \right)dy} .\]
The washer method for finding the volume of a solid generated by revolving about the y-axis.
Figure 4.

Volume of a Solid of Revolution for a Parametric Curve

If a bounding curve is defined in parametric form by the equations \(x = x\left( t \right),\) \(y = y\left( t \right),\) where the parameter \(t\) varies from \(\alpha\) to \(\beta,\) then the volume of the solid generated by revolving the curve about the \(x-\)axis is given by

\[{V_x} = \pi \int\limits_\alpha ^\beta {{y^2}\left( t \right)\frac{{dx}}{{dt}}dt} .\]

Respectively, when the curve is rotated about the \(y-\)axis, the volume of the solid of revolution is equal

\[{V_y} = \pi \int\limits_\alpha ^\beta {{x^2}\left( t \right)\frac{{dy}}{{dt}}dt} .\]

Volume of a Solid of Revolution for a Polar Curve

There are many curves that are given by a polar equation \(r = r\left( \theta \right).\) To convert from polar coordinates \(\left( {r,\theta } \right)\) to Cartesian coordinates \(\left( {x,y} \right),\) we use the known formulas

\[x = r\left( \theta \right)\cos \theta ,\;\; y = r\left( \theta \right)\sin \theta .\]

So we come to the parametric form of the curve considered in the previous section.

It is important to keep in mind that the radius vector \(r\) also depends on the parameter \(\theta.\) Therefore, the derivatives \(\frac{{dx}}{{dt}}\) and \(\frac{{dy}}{{dt}}\) are written as

\[\frac{{dx}}{{dt}} = \frac{{d\left( {r\left( \theta \right)\cos \theta } \right)}}{{dt}} = \frac{{d\left( {r\left( \theta \right)} \right)}}{{dt}}\cos \theta - r\left( \theta \right)\sin \theta ,\]
\[\frac{{dy}}{{dt}} = \frac{{d\left( {r\left( \theta \right)\sin \theta } \right)}}{{dt}} = \frac{{d\left( {r\left( \theta \right)} \right)}}{{dt}}\sin \theta + r\left( \theta \right)\cos \theta .\]

Solved Problems

Click or tap a problem to see the solution.

Example 1

Using the disk method, calculate the volume of the right circular cone of height \(H\) and base radius \(R.\)

Example 2

Find the volume of the solid obtained by rotating the sine function between \(x = 0\) and \(x = \pi\) about the \(x-\)axis.

Example 3

Calculate the volume of the solid obtained by rotating the region bounded by the parabola \(y = {x^2}\) and the square root function \(y = \sqrt x\) around the \(x-\)axis.

Example 4

Find the volume of the solid obtained by rotating the region bounded by two parabolas \[y = {x^2} + 1, y = 3 - {x^2}\] about the \(x-\)axis.

Example 1.

Using the disk method, calculate the volume of the right circular cone of height \(H\) and base radius \(R.\)

Solution.

The slant height of the cone is defined by the equation:

\[x = R - \frac{R}{H}y.\]
Finding the volume of the right circular cone using the disk method.
Figure 5.

Hence, the volume of the cone is given by

\[V = \pi \int\limits_0^H {{{\left[ {x\left( y \right)} \right]}^2}dy} = \pi \int\limits_0^H {{{\left[ {R - \frac{R}{H}y} \right]}^2}dy} = \pi {R^2}\int\limits_0^H {\left( {1 - \frac{{2y}}{H} + \frac{{{y^2}}}{{{H^2}}}} \right)dy} = \pi {R^2}\left. {\left( {H - \frac{{{y^2}}}{H} + \frac{{{y^3}}}{{3{H^2}}}} \right)} \right|_0^H = \pi {R^2}\left( {\cancel{H} -\cancel{H} + \frac{H}{3}} \right) = \frac{{\pi {R^2}H}}{3}.\]

Example 2.

Find the volume of the solid obtained by rotating the sine function between \(x = 0\) and \(x = \pi\) about the \(x-\)axis.

Solution.

Solid obtained by rotating the sine half-wave.
Figure 6.

By the disk method,

\[V = \pi \int\limits_0^\pi {{{\left[ {\sin x} \right]}^2}dx} = \frac{\pi }{2}\int\limits_0^\pi {\left( {1 - \cos 2x} \right)dx} = \frac{\pi }{2}\left. {\left( {x - \frac{{\sin 2x}}{2}} \right)} \right|_0^\pi = \frac{\pi }{2}\left[ {\left( {\pi - 0} \right) - \left( {0 - 0} \right)} \right] = \frac{{{\pi ^2}}}{2}.\]

Example 3.

Calculate the volume of the solid obtained by rotating the region bounded by the parabola \(y = {x^2}\) and the square root function \(y = \sqrt x\) around the \(x-\)axis.

Solution.

Solid formed by rotating the region bounded by parabola y=x^2 and the square root function y=sqrt(x) around the x-axis.
Figure 7.

Both curves intersect at the points \(x = 0\) and \(x = 1.\) Using the washer method, we have

\[V = \pi \int\limits_0^1 {\left( {{{\left[ {\sqrt x } \right]}^2} - {{\left[ {{x^2}} \right]}^2}} \right)dx} = \pi \int\limits_0^1 {\left( {x - {x^4}} \right)dx} = \pi \left. {\left( {\frac{{{x^2}}}{2} - \frac{{{x^5}}}{5}} \right)} \right|_0^1 = \pi \left( {\frac{1}{2} - \frac{1}{5}} \right) = \frac{{3\pi }}{{10}}.\]

Example 4.

Find the volume of the solid obtained by rotating the region bounded by two parabolas \[y = {x^2} + 1, y = 3 - {x^2}\] about the \(x-\)axis.

Solution.

First we determine the boundaries \(a\) and \(b:\)

\[{x^2} + 1 = 3 - {x^2},\;\; \Rightarrow 2{x^2} = 2,\;\; \Rightarrow {x^2} = 1,\;\;\Rightarrow x_{1,2} = \pm 1.\]

Hence the limits of integration are \(a = -1,\) \(b = 1.\) We sketch the bounding region and the solid of revolution:

Solid formed by revolving the region bounded by two parabolas.
Figure 8.

Using the washer method, we find the volume of the solid:

\[V = \pi \int\limits_a^b {\left( {{{\left[ {f\left( x \right)} \right]}^2} - {{\left[ {g\left( x \right)} \right]}^2}} \right)dx} = \pi \int\limits_{ - 1}^1 {\left( {{{\left( {3 - {x^2}} \right)}^2} - {{\left( {{x^2} + 1} \right)}^2}} \right)dx} = \pi \int\limits_{ - 1}^1 {\left( {{{\left[ {3 - {x^2}} \right]}^2} - {{\left[ {{x^2} + 1} \right]}^2}} \right)dx} = \pi \int\limits_{ - 1}^1 {\left( {8 - 8{x^2}} \right)dx} = 8\pi \int\limits_{ - 1}^1 {\left( {1 - {x^2}} \right)dx} = 8\pi \left. {\left( {x - \frac{{{x^3}}}{3}} \right)} \right|_{ - 1}^1 = 8\pi \left[ {\left( {1 - \frac{1}{3}} \right) - \left( { - 1 + \frac{1}{3}} \right)} \right] = 8\pi \cdot \frac{4}{3} = \frac{{32\pi }}{3}.\]

See more problems on Page 2.

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