Calculus

Applications of Integrals

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Volume of a Solid with a Known Cross Section

Solved Problems

Example 5.

Find the volume of a solid if the base of the solid is the circle given by the equation \[{x^2} + {y^2} = 1,\] and every perpendicular cross section is a square.

Solution.

A solid whose base is the circle x^2+y^2=1 and the cross section is a square.
Figure 6.

An arbitrary cross section at a point \(x\) has the side \(a\) equal to

\[a = 2y = 2\sqrt {1 - {x^2}} .\]

Hence, the area of the cross section is

\[A\left( x \right) = {a^2} = 4\left( {1 - {x^2}} \right).\]

Calculate the volume of the solid:

\[V = \int\limits_{ - 1}^1 {A\left( x \right)dx} = \int\limits_{ - 1}^1 {4\left( {1 - {x^2}} \right)dx} = 4\left. {\left( {x - \frac{{{x^3}}}{3}} \right)} \right|_{ - 1}^1 = 4\left[ {\left( {1 - \frac{{{1^3}}}{3}} \right) - \left( { - 1 - \frac{{{{\left( { - 1} \right)}^3}}}{3}} \right)} \right] = 4\left[ {\frac{2}{3} - \left( { - \frac{2}{3}} \right)} \right] = \frac{{16}}{3}.\]

Example 6.

Find the volume of the frustum of a cone if its bases are ellipses with the semi-axes \(A, B,\) and \(a, b\), and the altitute is equal to \(H.\)

Solution.

Frustum of a cone with the ellipses as bases
Figure 7.

The volume of the frustum of the cone is given by the integral

\[V = \int\limits_0^H {A\left( x \right)dx} ,\]

where \({A\left( x \right)}\) is the cross-sectional area at a point \(x.\)

The lengths of the major and minor axes linearly change from \(a, b\) to \(A, B,\) and at the point \(x\) they are determined by the following expressions:

\[\text{major axis:}\;\;a + \left( {A - a} \right)\frac{x}{H}\;\; \text{minor axis:}\;\;b + \left( {B - b} \right)\frac{x}{H}.\]

Let's now calculate the area of the cross section:

\[A\left( x \right) = \pi \left( {a + \left( {A - a} \right)\frac{x}{H}} \right)\left( {b + \left( {B - b} \right)\frac{x}{H}} \right) = \pi \left[ {ab + b\left( {A - a} \right)\frac{x}{H} + a\left( {B - b} \right)\frac{x}{H} + \left( {A - a} \right)\left( {B - b} \right){{\left( {\frac{x}{H}} \right)}^2}} \right] = \pi \left[ {ab + \left( {bA + aB - 2ab} \right)\frac{x}{H} + \left( {AB - aB - bA + ab} \right){{\left( {\frac{x}{H}} \right)}^2}} \right].\]

Then the volume is given by

\[V = \int\limits_0^H {A\left( x \right)dx} = \pi \left[ {abH + \left( {bA + ab - 2ab} \right)\frac{H}{2} + \left( {AB - aB - bA + ab} \right)\frac{H}{3}} \right] = \pi \left[ {\cancel{abH} + \frac{{bAH}}{2} + \frac{{aBH}}{2} - \cancel{abH} + \frac{{ABH}}{3} - \frac{{aBH}}{3} - \frac{{bAH}}{3} + \frac{{abH}}{3}} \right] = \pi \left[ {\frac{{bAH}}{6} + \frac{{aBH}}{6} + \frac{{ABH}}{3} + \frac{{abH}}{3}} \right] = \frac{{\pi H}}{6}\left[ {bA + aB + 2AB + 2ab} \right] = \frac{{\pi H}}{6}\left[ {\left( {2A + a} \right)B + \left( {A + 2a} \right)b} \right].\]

Example 7.

Calculate the volume of a wedge given the bottom sides \(a, b,\) the top side \(c,\) and the altitude \(H.\)

Solution.

Regular wedge with the bottom sides a,b, the top side c and the height h.
Figure 8.

Consider an arbitrary cross section at a height \(x.\) This cross section is a rectangle with the sides \(m\) and \(n.\) It is easy to see that

\[m = c + \left( {a - c} \right)\frac{x}{h},\;\; n = \frac{{bx}}{h}.\]

Then the cross-sectional area \(A\left( x \right)\) is written as

\[A\left( x \right) = mn = \left( {c + \left( {a - c} \right)\frac{x}{h}} \right)\frac{{bx}}{h} = \frac{{bcx}}{h} + \frac{{ab{x^2}}}{{{h^2}}} - \frac{{bc{x^2}}}{{{h^2}}} = \frac{{bcx}}{h} + \frac{{b\left( {a - c} \right){x^2}}}{{{h^2}}}.\]

We find the volume of the wedge by integration:

\[V = \int\limits_0^h {A\left( x \right)dx} = \int\limits_0^h {\left( {\frac{{bcx}}{h} + \frac{{b\left( {a - c} \right){x^2}}}{{{h^2}}}} \right)dx} = \left. {\frac{{bc{x^2}}}{{2h}} + \frac{{b\left( {a - c} \right){x^3}}}{{3{h^2}}}} \right|_o^h = \frac{{bch}}{2} + \frac{{b\left( {a - c} \right)h}}{3} = \frac{{bch}}{2} + \frac{{abh}}{3} - \frac{{bch}}{3} = \frac{{bch}}{6} + \frac{{abh}}{3} = \frac{{bh}}{6}\left( {2a + c} \right).\]

Example 8.

Find the volume of a regular tetrahedron with the edge \(a.\)

Solution.

Regular tetrahedron with the edge a
Figure 9.

The base of the tetrahedron is an equilateral triangle. Calculate the altitude of the base \(CE.\) By Pythagorean theorem,

\[{{CE}^2} = {{BC}^2} -{{EB}^2},\;\; \Rightarrow CE = \sqrt {{BC^2} - {EB^2}} = \sqrt {{a^2} - {{\left( {\frac{a}{2}} \right)}^2}} = \frac{{\sqrt 3 a}}{2}.\]

Hence

\[CO = \frac{2}{3}CE = \frac{{\sqrt 3 a}}{3}.\]

We express the altitude of the tetrahedron \(h\) in terms of \(a:\)

\[h = DO = \sqrt {{DC^2} - {CO^2}} = \sqrt {{a^2} - {{\left( {\frac{{\sqrt 3 a}}{3}} \right)}^2}} = \sqrt {\frac{{2{a^2}}}{3}} = \frac{{\sqrt 2 a}}{{\sqrt 3 }}.\]

The cross-sectional area \(A\left( x \right)\) is written in the form

\[A\left( x \right) = \frac{1}{2}{m^2}\sin 60^{\circ} = \frac{{\sqrt 3 {m^2}}}{4},\]

where \(m\) is the side of the equilateral triangle in the cross section.

It follows from the similarity that

\[m = \frac{{ax}}{h} = \frac{{\sqrt 3 \cancel{a}x}}{{\sqrt 2 \cancel{a}}} = \frac{{\sqrt 3 x}}{{\sqrt 2 }}.\]

Using integration, we find the volume:

\[V = \int\limits_0^h {A\left( x \right)dx} = \int\limits_0^h {\frac{{\sqrt 3 {m^2}}}{4}dx} = \int\limits_0^h {\frac{{\sqrt 3 }}{4}{{\left( {\frac{{\sqrt 3 x}}{{\sqrt 2 }}} \right)}^2}dx} = \int\limits_0^h {\frac{{3\sqrt 3 {x^2}}}{8}dx} = \frac{{3\sqrt 3 }}{8}\int\limits_0^h {{x^2}dx} = \frac{{3\sqrt 3 }}{8} \cdot \left. {\frac{{{x^3}}}{3}} \right|_0^h = \frac{{\sqrt 3 {h^3}}}{8} = \frac{{\sqrt 3 }}{8} \cdot {\left( {\frac{{\sqrt 2 a}}{{\sqrt 3 }}} \right)^3} = \frac{{2\cancel{\sqrt 3} \sqrt 2 {a^3}}}{{24\cancel{\sqrt 3} }} = \frac{{\sqrt 2 {a^3}}}{{12}}.\]

Example 9.

A wedge is cut out of a circular cylinder with radius \(R\) and height \(H\) by the plane passing through a diameter of the base (Figure \(10\)). Find the volume of the cylindrical wedge.

Solution.

A wedge cut out of a cylinder of radius R and height H by the plane passing through a diameter of the base.
Figure 10.

A cross section of the wedge perpendicular to the \(x-\)axis is a right triangle \(ABC.\) The leg of the triangle \(AB\) is given by

\[AB = y = \sqrt {{R^2} - {x^2}} ,\]

and the other leg \(BC\) is expressed in the form

\[BC = AB \cdot \tan \alpha = AB \cdot \frac{H}{R} = \frac{H}{R}\sqrt {{R^2} - {x^2}}.\]

Hence, the area of the cross section is written as

\[A\left( x \right) = \frac{{AB \cdot BC}}{2} = \frac{H}{R}{\left( {\sqrt {{R^2} - {x^2}} } \right)^2} = \frac{H}{R}\left( {{R^2} - {x^2}} \right).\]

Integrating yields

\[V = 2\int\limits_0^R {A\left( x \right)dx} = 2\int\limits_0^R {\frac{H}{{2R}}\left( {{R^2} - {x^2}} \right)dx} = \frac{H}{R}\int\limits_0^R {\left( {{R^2} - {x^2}} \right)dx} = \frac{H}{R}\left. {\left( {{R^2}x - \frac{{{x^3}}}{3}} \right)} \right|_0^R = \frac{H}{R} \cdot \frac{{2{R^3}}}{3} = \frac{{2{R^2}H}}{3},\]

so the volume of the wedge is

\[\frac{{\frac{{2{R^2}H}}{3}}}{{\pi {R^2}H}} = \frac{2}{{3\pi }} \approx 0.212\]

of the total volume of the cylinder. The result does not depend on \(R\) and \(H!\)

Example 10.

The axes of two circular cylinders with the same radius \(R\) intersect at right angles. Find the volume of the solid common to both these cylinders.

Solution.

The figure below shows \(\frac{1}{8}\) of the solid of intersection.

Two circular cylinders intersecting at right angles.
Figure 11.

Consider a cross section \(ABCD\) perpendicular to the \(x-\)axis at an arbitrary point \(x\). Due to symmetry, the cross section is a square with sides of length

\[BC = AD = y = \sqrt {{R^2} - {x^2}} ,\]
\[AB = CD = z = \sqrt {{R^2} - {x^2}}.\]

The cross-sectional area is expressed in terms of \(x\) as follows:

\[A\left( x \right) = {\left( {\sqrt {{R^2} - {x^2}} } \right)^2} = {R^2} - {x^2}.\]

Then the volume of the solid common to both the cylinders (bicylinder) is given by

\[V = 8\int\limits_0^R {A\left( x \right)dx} = 8\int\limits_0^R {\left( {{R^2} - {x^2}} \right)dx} = 8\left. {\left( {{R^2}x - \frac{{{x^3}}}{3}} \right)} \right|_0^R = 8 \cdot \frac{{2{R^3}}}{3} = \frac{{16{R^3}}}{3}.\]
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