Calculus

Applications of Integrals

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Pappus’s Theorem

Solved Problems

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Example 5

Find the centroid of a right triangle with legs a, b.

Example 6

Find the centroid of the region enclosed by a half-wave of the sine curve and the x-axis.

Example 7

A curve shown in Figure \(10\) is rotated about the \(y-\)axis. Find the area of the surface of revolution.

Example 8

A square of side \(a\) is rotated about an axis lying in its plane and passing through one of the vertices. The angle between the square side and positive direction of the axis of rotation is \(\alpha.\) What is the volume of the obtained solid of revolution?

Example 9

Find the centroid of the region bounded by the parabola \(y = 4 - {x^2}\) and the \(x-\)axis.

Example 10

A circular arc of radius \(R\) subtending the central angle \(2\alpha\) is rotated about the \(x-\)axis as shown in Figure \(13.\) Determine the centroid of the arc.

Example 5.

Find the centroid of a right triangle with legs \(a, b.\)

Solution.

Centroid of a right triangle with legs a,b.
Figure 8.

To determine the coordinates of the centroid, we will use the \(2\text{nd}\) theorem of Pappus.

Suppose first that the triangle is rotated about the \(y-\)axis. The volume of the obtained cone is given by

\[{V_y} = \frac{{\pi {a^2}b}}{3}.\]

The area of the triangle is

\[A = \frac{{ab}}{2}.\]

Then, by the Pappus's theorem,

\[{V_y} = 2\pi \bar xA,\;\; \Rightarrow \bar x = \frac{{V_y}}{{2\pi A}} = \frac{{\frac{{\pi {a^2}b}}{3}}}{{2\pi \cdot \frac{{ab}}{2}}} = \frac{a}{3}\]

Let the triangle rotate now about the \(x-\)axis. Similarly, we find the volume

\[{V_x} = \frac{{\pi a{b^2}}}{3}\]

and the \(\bar y-\)coordinate of the centroid:

\[{V_x} = 2\pi \bar yA,\;\; \Rightarrow \bar y = \frac{{{V_x}}}{{2\pi A}} = \frac{{\frac{{\pi a{b^2}}}{3}}}{{2\pi \cdot \frac{{ab}}{2}}} = \frac{b}{3}\]

Thus, the centroid of the triangle is located at the point

\[G\left( {\bar x,\bar y} \right) = G\left( {\frac{a}{3},\frac{b}{3}} \right),\]

which is the point of intersection of its medians.

Example 6.

Find the centroid of the region enclosed by a half-wave of the sine curve and the \(x-\)axis.

Solution.

Centroid of the region bounded by a half-wave of the sine curve and the x-axis.
Figure 9.

Let the point \(G\left( {\bar x,\bar y} \right)\) denote the centroid of the figure. By symmetry, \(\bar x = \frac{\pi }{2},\) so we need to calculate only the coordinate \(\bar y = m.\)

Using the disk method, we find the volume of the solid of revolution:

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

The area under the sine curve is

\[A = \int\limits_a^b {f\left( x \right)dx} = \int\limits_0^\pi {\sin xdx} = \left. {\left( { - \cos x} \right)} \right|_0^\pi = - \cos \pi + \cos 0 = 2.\]

The \(2\text{nd}\) theorem of Pappus states that

\[V = Ad = 2\pi mA.\]

Hence,

\[m = \frac{V}{{2\pi A}} = \frac{{\frac{{{\pi ^2}}}{2}}}{{4\pi }} = \frac{\pi }{8} \approx 0.39\]

Thus, the centroid of the region has the coordinates

\[G\left( {\bar x,\bar y} \right) = G\left( {\frac{\pi }{2},\frac{\pi }{8}} \right).\]

Example 7.

A curve shown in Figure \(10\) is rotated about the \(y-\)axis. Find the area of the surface of revolution.

Solution.

A surface of revolution obtained by rotating a curve in the shape of a number 5.
Figure 10.

We consider separately three sections of the curve and compute their centroids.

  1. Horizontal line segment \(AB.\)
    The length is \({L_{AB}} = 3.\) The centroid is located at the point \({G_{AB}} = \left( {3.5,10} \right);\)
  2. Vertical line segment \(BC.\)
    The length is \({L_{BC}} = 2.\) The centroid is located at the point \({G_{BC}} = \left( {2,9} \right);\)
  3. Semicircular arc \(CD.\)
    The length is \({L_{CD}} = \pi R = 3\pi.\) The centroid is located at the point \({G_{CD}} = \left( {\bar x_{CD},\bar y_{CD}} \right),\) where
    \[{{\bar x}_{CD}} = 2 + \frac{{2R}}{\pi } = 2 + \frac{6}{\pi };\;\; {{\bar y}_{CD}} = 5.\]

Calculate the \(\bar x-\)coordinate of the centroid \(G\) of the whole curve:

\[\bar x = \frac{{{{\bar x}_{AB}}{L_{AB}} + {{\bar x}_{BC}}{L_{BC}} + {{\bar x}_{CD}}{L_{CD}}}}{L},\]

where \(L = {L_{AB}} + {L_{BC}} + {L_{CD}}\) is the total length of the curve.

By the \(1\text{st}\) theorem of Pappus, the surface area is given by

\[A = Ld = 2\pi mL,\]

where \(d\) is the path traversed by the centroid of the curve in one turn and \(m = \bar x\) is the distance from the centroid to the \(y-\)axis.

Hence,

\[A = 2\pi \cdot \frac{{{{\bar x}_{AB}}{L_{AB}} + {{\bar x}_{BC}}{L_{BC}} + {{\bar x}_{CD}}{L_{CD}}}}{\cancel{L}} \cdot \cancel{L} = 2\pi \left( {{{\bar x}_{AB}}{L_{AB}} + {{\bar x}_{BC}}{L_{BC}} + {{\bar x}_{CD}}{L_{CD}}} \right) = 2\pi \left[ {3.5 \cdot 3 + 2 \cdot 2 + \left( {2 + \frac{6}{\pi }} \right) \cdot 3\pi } \right] = 61\pi + 12{\pi ^2} \approx 310.\]

Example 8.

A square of side \(a\) is rotated about an axis lying in its plane and passing through one of the vertices. The angle between the square side and positive direction of the axis of rotation is \(\alpha.\) What is the volume of the obtained solid of revolution?

Solution.

A square rotating about one of the vertices.
Figure 11.

The half of the diagonal of the square \(AG\) has the length

\[AG = \frac{{a\sqrt 2 }}{2}.\]

The angle \(\beta = \angle KGA\) is expressed in terms of \(\alpha\) as follows:

\[\beta = 45 ^{\circ}- \alpha .\]

Hence, the distance \(m\) from the centroid \(G\) to the axis of revolution is given by

\[m = KG = AG\cos \beta = \frac{{a\sqrt 2 }}{2}\cos \left( {45^{\circ} - \alpha } \right).\]

Using the cosine subtraction identity

\[\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B,\]

we write the distance \(m\) in the form

\[m = \frac{{a\sqrt 2 }}{2}\cos \left( {45^{\circ} - \alpha } \right) = \frac{{a\sqrt 2 }}{2}\left( {\cos 45^{\circ}\cos \alpha + \sin 45^{\circ}\sin \alpha } \right) = \frac{{a\sqrt 2 }}{2}\left( {\frac{{\sqrt 2 }}{2}\cos \alpha + \frac{{\sqrt 2 }}{2}\sin \alpha } \right) = \frac{a}{2}\left( {\cos \alpha + \sin \alpha } \right),\]

so the path \(d\) traversed by the centroid \(G\) of the square is given by

\[d = 2\pi m = \pi a\left( {\cos \alpha + \sin \alpha } \right).\]

Applying the \(2\text{nd}\) theorem of Pappus, we find the volume of the solid of revolution:

\[V = Ad = {a^2} \cdot \pi a\left( {\cos \alpha + \sin \alpha } \right) = \pi {a^3}\left( {\cos \alpha + \sin \alpha } \right).\]

Considering the volume as a function of angle \(\alpha\) we can determine its largest value:

\[V\left( \alpha \right) = \pi {a^3}\left( {\cos \alpha + \sin \alpha } \right),\]
\[V^\prime\left( \alpha \right) = \pi {a^3}\left( {\cos \alpha + \sin \alpha } \right)^\prime = \pi {a^3}\left( { - \sin \alpha + \cos \alpha } \right).\]
\[V^\prime\left( \alpha \right) = 0,\;\; \Rightarrow \pi {a^3}\left( {\cos \alpha - \sin \alpha } \right) = 0,\;\; \Rightarrow \tan \alpha = 1,\;\; \Rightarrow \alpha = \frac{\pi }{4}.\]

By the second derivative test,

\[V^{\prime\prime}\left( \alpha \right) = \pi {a^3}\left( { - \sin \alpha + \cos \alpha } \right)^{\prime} = - \pi {a^3}\left( {\cos \alpha + \sin \alpha } \right),\]

and, consequently,

\[V^{\prime\prime}\left( {\frac{\pi }{4}} \right) = - \pi {a^3}\left( {\cos \frac{\pi }{4} + \sin \frac{\pi }{4}} \right) = - \pi {a^3}\sqrt 2 \lt 0.\]

So the volume has the maximum at \(\alpha = \frac{\pi }{4}:\)

\[{V_{\max }} = V\left( {\frac{\pi }{4}} \right) = \pi {a^3}\left( {\cos \frac{\pi }{4} + \sin \frac{\pi }{4}} \right) = \pi {a^3}\sqrt 2 .\]

Example 9.

Find the centroid of the region bounded by the parabola \(y = 4 - {x^2}\) and the \(x-\)axis.

Solution.

Centroid of the region bounded by the parabola y=4-x^2 and the x-axis.
Figure 12.

Using the Pappus's theorem for volume, we have

\[V = Ad = 2\pi mA,\]

where \(A\) is the area of the region and \(m\) is the \(\bar y-\)coordinate of the centroid \(G\left( {\bar x,\bar y} \right).\)

We compute the volume of the solid of revolution using the disk method:

\[V = \pi \int\limits_a^b {{f^2}\left( x \right)dx} = \pi \int\limits_{ - 2}^2 {{{\left( {4 - {x^2}} \right)}^2}dx} = 2\pi \int\limits_0^2 {{{\left( {4 - {x^2}} \right)}^2}dx} = 2\pi \int\limits_0^2 {\left( {16 - 8{x^2} + {x^4}} \right)dx} = 2\pi \left. {\left( {16x - \frac{{8{x^3}}}{3} + \frac{{{x^5}}}{5}} \right)} \right|_0^2 = 2\pi \left( {32 - \frac{{64}}{3} + \frac{{32}}{5}} \right) = \frac{{512\pi }}{{15}}.\]

The area of the region is

\[A = \int\limits_a^b {f\left( x \right)dx} = \int\limits_{ - 2}^2 {\left( {4 - {x^2}} \right)dx} = 2\int\limits_0^2 {\left( {4 - {x^2}} \right)dx} = 2\left. {\left( {4x - \frac{{{x^3}}}{3}} \right)} \right|_0^2 = 2\left( {8 - \frac{8}{3}} \right) = \frac{{32}}{3}.\]

Then the \(\bar y-\)coordinate of the centroid is given by

\[m = \frac{V}{{2\pi A}} = \frac{{\frac{{512\pi }}{{15}}}}{{2\pi \cdot \frac{{32}}{3}}} = \frac{8}{5}.\]

Due to symmetry of the region, the \(\bar x-\)coordinate is equal to \(0,\) so the final answer is

\[G\left( {\bar x,\bar y} \right) = G\left( {0,\frac{8}{5}} \right).\]

Example 10.

A circular arc of radius \(R\) subtending the central angle \(2\alpha\) is rotated about the \(x-\)axis as shown in Figure \(13.\) Determine the centroid of the arc.

Solution.

Centroid of a circular arc of radius R subtending the central angle 2alpha.
Figure 13.

By symmetry, the centroid \(G\) is located on the \(y-\)axis, so its coordinates are

\[G\left( {\bar x,\bar y} \right) = G\left( {0,m} \right),\]

where \(m = \bar y\) is the distance from the centroid to the axis of rotation that we're going to find.

When the arc is rotated it forms a spherical segment. The surface area \(A\) of the spherical segment is given by

\[A = 2\pi Rh,\]

where \(h\) is the distance between the parallel planes cutting the sphere.

Since \(h = 2R\sin\alpha,\) we can write

\[A = 2\pi R \cdot 2R\sin \alpha = 4\pi {R^2}\sin \alpha .\]

From the other side, by the \(1\text{st}\) theorem of Pappus, we have

\[A = dL = 2\pi mL,\]

where \(d = 2\pi m\) is the path traversed by the centroid in one turn and \(L = 2\alpha R\) is the length of the arc.

From here it follows that

\[m = \bar y = \frac{A}{{2\pi L}} = \frac{{4\pi {R^2}\sin \alpha }}{{2\pi \cdot 2\alpha R}} = \frac{{R\sin \alpha }}{\alpha }.\]

We can highlight some particular cases:

  1. If \(\alpha = 0,\) then
    \[m\left( {\alpha = 0} \right) = \lim\limits_{\alpha \to 0} \frac{{R\sin \alpha }}{\alpha } = R\,\underbrace {\lim\limits_{\alpha \to 0} \frac{{\sin \alpha }}{\alpha }}_1 = R;\]
  2. If \(\alpha = \pi,\) then
    \[m\left( {\alpha = \pi } \right) = \frac{{R\sin \pi }}{\pi } = 0;\]
  3. If \({\alpha = \frac{\pi }{2}},\) then
    \[m\left( {\alpha = \frac{\pi }{2}} \right) = \frac{{2R\sin \frac{\pi }{2}}}{\pi } = \frac{{2R}}{\pi }.\]
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