Calculus

Applications of Integrals

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Volume of a Solid of Revolution: Cylindrical Shells

Solved Problems

Example 5.

Calculate the volume of a sphere of radius \(R\) using the shell integration.

Solution.

Sphere obtained by rotating a semi-disk of radius R.
Figure 6.

We consider the region which occupies the part of the disk \({x^2} + {y^2} = {R^2}\) in the first quadrant. Taking the integral \(2\pi \int\limits_0^R {x\sqrt {{R^2} - {x^2}} dx} \) and multiplying the result by two, we find the volume of the sphere.

It's convenient to change the variable. Let \({R^2} - {x^2} = {t^2},\) so \(xdx=-tdt.\) When \(x = 0,\) then \(t = R,\) and when \(x = R,\) then \(t = 0.\) Hence, the integral becomes

\[V = 4\pi \int\limits_0^R {x\sqrt {{R^2} - {x^2}} dx} = 4\pi \int\limits_R^0 {t\left( { - tdt} \right)} = 4\pi \int\limits_0^R {{t^2}dt} = \left. {\frac{{4\pi {t^3}}}{3}} \right|_0^R = \frac{{4\pi {R^3}}}{3}.\]

Example 6.

Find the volume of the solid obtained by rotating about the \(y-\)axis the region bounded by the curve \(y = {e^{ - x}},\) where \(0 \le x \lt \infty ,\) and the horizontal line \(y = 0.\)

Solution.

Solid obtained by rotating the infinite region bounded by the curve y=exp(-x), the x-axis, and the y-axis.
Figure 7.

The volume of the solid is expressed by the improper integral

\[V = 2\pi \int\limits_0^\infty {x{e^{ - x}}dx} .\]

To take the integral we use integration by parts:

\[V = 2\pi \int\limits_0^\infty {x{e^{ - x}}dx} = \left[ {\begin{array}{*{20}{l}} {u = x}\\ {dv = {e^{ - x}}dx}\\ {du = dx}\\ {v = - {e^{ - x}}} \end{array}} \right] = 2\pi \left\{ {\left. {\left[ { - x{e^{ - x}}} \right]} \right|_0^\infty - \int\limits_0^\infty {\left( { - {e^{ - x}}} \right)dx} } \right\} = 2\pi \left\{ {\left. {\left[ { - x{e^{ - x}}} \right]} \right|_0^\infty + \int\limits_0^\infty {{e^{ - x}}dx} } \right\} = 2\pi \left\{ {\left. {\left[ { - x{e^{ - x}}} \right]} \right|_0^\infty - \left. {\left[ {{e^{ - x}}} \right]} \right|_0^\infty } \right\} = - 2\pi \left. {\left[ {{e^{ - x}}\left( {x + 1} \right)} \right]} \right|_0^\infty .\]

We replace the infinite upper limit with a finite value \(b\) and then take the limit as \(b \to \infty.\)

\[V = - 2\pi \lim\limits_{b \to \infty } \left. {\left[ {{e^{ - x}}\left( {x + 1} \right)} \right]} \right|_0^b = - 2\pi \lim\limits_{b \to \infty } \left[ {{e^{ - b}}\left( {b + 1} \right) - {e^0}\left( {0 + 1} \right)} \right] = 2\pi \lim\limits_{b \to \infty } \left[ {1 - \frac{{b + 1}}{{{e^b}}}} \right].\]

By L'Hopital's Rule,

\[\lim \limits_{b \to \infty } \frac{{b + 1}}{{{e^b}}} = \lim \limits_{b \to \infty } \frac{{\left( {b + 1} \right)'}}{{\left( {{e^b}} \right)'}} = \lim \limits_{b \to \infty } \frac{1}{{{e^b}}} = 0.\]

So, the volume of the solid is \(V = 2\pi .\)

Example 7.

Find the volume of the solid formed by rotating about the line \(x = -1\) the region bounded by the parabola \(y = {x^2}\) and the lines \(y = 0,\) \(x = 1.\)

Solution.

Solid formed by rotating about the line x=-1 the region bounded by the curve y=x^2 and the lines y=0, x=1.
Figure 8.

The curve is rotating about the line \(x = -1,\) which is on the left from the interval of integration \(\left[ {0,1} \right].\) Therefore, we use the following formula for calculating the volume:

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

where \(h = -1,\) \(a = 0,\) \(b = 1.\) So, we have

\[V = 2\pi \int\limits_0^1 {\left( {x + 1} \right){x^2}dx} = 2\pi \int\limits_0^1 {\left( {{x^3} + {x^2}} \right)dx} = 2\pi \left. {\left[ {\frac{{{x^4}}}{4} + \frac{{{x^3}}}{3}} \right]} \right|_0^1 = 2\pi \left( {\frac{1}{4} + \frac{1}{3}} \right) = \frac{{7\pi }}{6}.\]

Example 8.

The region bounded by curves \[y = {x^2}, y = \sqrt x \] rotates around the \(y-\)axis. Find the volume of the solid of revolution.

Solution.

Solid obtained by rotating about the y-axis the region bounded by the parabola y=x^2 and the square root function y=sqrt(x).
Figure 9.

Obviously, both curves intersect at the points \(x = 0\) and \(x = 1,\) so using the shell method we will integrate from \(0\) to \(1\).

The region here is bounded by two curves. Therefore the integration formula is written in the form similar to the washer method:

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

Substituting \(a = 0,\) \(b= 1,\) \(f\left( x \right) = \sqrt x,\) \(g\left( x \right) = {x^2},\) we obtain:

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

Compare the result with Example \(3\) on the page Volume of a Solid of Revolution: Disks and Washers.

Example 9.

The region is bounded by the hyperbola \(y = \frac{1}{x}\) and the line \(y = \frac{5}{2} - x.\) Find the volume of the solid formed by rotating the region about the \(y-\)axis.

Solution.

Solid formed by rotating about the y-axis the region bounded by the hyperbola y=1/x and the line y=5/2-x.
Figure 10.

First we determine the points of intersection of both curves:

\[\frac{1}{x} = \frac{5}{2} - x,\;\; \Rightarrow {x^2} - \frac{5}{2}x + 1 = 0,\;\; \Rightarrow 2{x^2} - 5x + 2 = 0,\;\; \Rightarrow D = {\left( { - 5} \right)^2} - 4 \cdot 2 \cdot 2 = 9,\;\; \Rightarrow {x_{1,2}} = \frac{{5 \pm \sqrt 9 }}{4} = \frac{1}{2},\,2.\]

We use the shell method for finding the volume. As the region is bounded by two curves, the integration formula is given by

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

Here \(a = \frac{1}{2},\) \(b = 2,\) \(f\left( x \right) = \frac{5}{2} - x,\) \(g\left( x \right) = \frac{1}{x}.\) So the volume of the solid is

\[V = 2\pi \int\limits_{\frac{1}{2}}^2 {x\left[ {\frac{5}{2} - x - \frac{1}{x}} \right]dx} = 2\pi \int\limits_{\frac{1}{2}}^2 {\left[ {\frac{{5x}}{2} - {x^2} - 1} \right]dx} = 2\pi \left. {\left[ {\frac{{5{x^2}}}{4} - \frac{{{x^3}}}{3} - x} \right]} \right|_{\frac{1}{2}}^2 = 2\pi \left[ {\left( {5 - \frac{8}{3} - 2} \right) - \left( {\frac{5}{{16}} - \frac{1}{{24}} - \frac{1}{2}} \right)} \right] = 2\pi \left[ {\frac{7}{2} - \frac{8}{3} - \frac{5}{{16}} + \frac{1}{{24}}} \right] = 2\pi \cdot \frac{9}{{16}} = \frac{{9\pi }}{8}.\]

Example 10.

Find the volume of the solid obtained by rotating about the line \(y = 2\) the region bounded by the curve \[x = 1 + {y^2}\] and the lines \(y = 0,\) \(y = 1.\)

Solution.

Solid formed by rotating about the axis y=2 the region bounded by the parabola x=y^2+1 and the lines y=0, y=1.
Figure 11.

The given figure rotates around the horizontal axis \(y = m = 2.\) To determine the volume of the solid of revolution, we apply the shell integration formula in the form

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

In this case, \(m = 2,\) \(c = 0,\) \(d = 1,\) \(x\left( y \right) = 1 + {y^2}.\) This yields

\[V = 2\pi \int\limits_0^1 {\left( {2 - y} \right)\left( {1 + {y^2}} \right)dy} = 2\pi \int\limits_0^1 {\left( {2 - y + 2{y^2} - {y^3}} \right)dy} = 2\pi \left. {\left[ {2y - \frac{{{y^2}}}{2} + \frac{{2{y^3}}}{3} - \frac{{{y^4}}}{4}} \right]} \right|_0^1 = 2\pi \left[ {2 - \frac{1}{2} + \frac{2}{3} - \frac{1}{4}} \right] = \frac{{23\pi }}{6}.\]
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