Calculus

Applications of Integrals

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Volume of a Solid with a Known Cross Section

In this topic, we will learn how to find the volume of a solid object that has known cross sections.

We consider solids whose cross sections are common shapes such as triangles, squares, rectangles, trapezoids, and semicircles.

Definition: Volume of a Solid Using Integration

Let S be a solid and suppose that the area of the cross section in the plane perpendicular to the x-axis is A (x) for a ≤ x ≤ b.

Volume of a solid with known cross section
Figure 1.

Then the volume of the solid from \(x = a\) to \(x = b\) is given by the cross-section formula

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

Similarly, if the cross section is perpendicular to the \(y-\)axis and its area is defined by the function \(A\left( y \right),\) then the volume of the solid from \(y = c\) to \(y = d\) is given by

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

Steps for Finding the Volume of a Solid with a Known Cross Section

  1. Sketch the base of the solid and a typical cross section.
  2. Express the area of the cross section \(A\left( x \right)\) as a function of \(x.\)
  3. Determine the limits of integration.
  4. Evaluate the definite integral
    \[V = \int\limits_a^b {A\left( x \right)dx}.\]

Solved Problems

Example 1.

The solid has a base lying in the first quadrant of the \(xy-\)plane and bounded by the lines \(y = x,\) \(x = 1,\) \(y = 0.\) Every planar section perpendicular to the \(x-\)axis is a semicircle. Find the volume of the solid.

Solution.

Solid whose base is a right triangle and cross sections are semicircles.
Figure 2.

The diameter of the semicircle at a point \(x\) is \(d=y=x.\) Hence, the area of the cross section is

\[A(x) = \frac{{\pi {d^2}}}{8} = \frac{{\pi {x^2}}}{8}.\]

Integration yields the following result:

\[V = \int\limits_0^1 {A\left( x \right)dx} = \int\limits_0^1 {\frac{{\pi {x^2}}}{8}dx} = \frac{\pi }{8}\int\limits_0^1 {{x^2}dx} = \frac{\pi }{8} \cdot \left. {\frac{{{x^3}}}{3}} \right|_0^1 = \frac{\pi }{{24}}.\]

Example 2.

Find the volume of a solid bounded by the elliptic paraboloid \[z = \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}\] and the plane \(z = 1.\)

Solution.

Solid bounded by the elliptic paraboloid z=x^2/a^2+y^2/b^2 and the plane z=1.
Figure 3.

Consider an arbitrary planar section perpendicular to the \(z-\)axis at a point \(z,\) where \(0 \lt z \le 1.\) The cross section is an ellipse defined by the equation

\[z = \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}},\;\; \Rightarrow \frac{{{x^2}}}{{{{\left( {a\sqrt z } \right)}^2}}} + \frac{{{y^2}}}{{{{\left( {b\sqrt z } \right)}^2}}} = 1.\]

The area of the cross section is

\[A\left( z \right) = \pi \cdot \left( {a\sqrt z } \right) \cdot \left( {b\sqrt z } \right) = \pi abz.\]

Then, by the cross-section formula,

\[V = \int\limits_0^1 {A\left( z \right)dz} = \int\limits_0^1 {\pi abzdz} = \pi ab\int\limits_0^1 {zdz} = \pi ab \cdot \left. {\frac{{{z^2}}}{2}} \right|_0^1 = \frac{{\pi ab}}{2}.\]

Example 3.

The base of a solid is bounded by the parabola \[y = 1 - {x^2}\] and the \(x-\)axis. Find the volume of the solid if the cross sections are equilateral triangles perpendicular to the \(x-\)axis.

Solution.

Solid with the base bounded by the parabola y=1-x^2 and cross sections formed by equilateral triangles.
Figure 4.

The area of the equilateral triangle at a point \(x\) is given by

\[A\left( x \right) = \frac{{{a^2}\sqrt 3 }}{4}.\]

As the side \(a\) is equal to \(1-{x^2},\) then

\[A\left( x \right) = \frac{{{a^2}\sqrt 3 }}{4} = \frac{{\sqrt 3 }}{4}{\left( {1 - {x^2}} \right)^2}.\]

The parabola \(y = 1 - {x^2}\) intersects the \(x-\)axis at the points \(x=-1,\) \(x = 1.\)

Compute the volume of the solid:

\[V = \int\limits_{ - 1}^1 {A\left( x \right)dx} = \int\limits_{ - 1}^1 {\frac{{\sqrt 3 }}{4}{{\left( {1 - {x^2}} \right)}^2}dx} = \frac{{\sqrt 3 }}{4}\int\limits_{ - 1}^1 {\left( {1 - 2{x^2} + {x^4}} \right)dx} = \frac{{\sqrt 3 }}{4}\left. {\left[ {x - \frac{{2{x^3}}}{3} + \frac{{{x^5}}}{5}} \right]} \right|_{ - 1}^1 = \frac{{\sqrt 3 }}{4}\left[ {\left( {1 - \frac{2}{3} + \frac{1}{5}} \right) - \left( { - 1 + \frac{2}{3} - \frac{1}{5}} \right)} \right] = \frac{{\sqrt 3 }}{2}\left( {1 - \frac{2}{3} + \frac{1}{5}} \right) = \frac{{4\sqrt 3 }}{{15}}.\]

Example 4.

Find the volume of a regular square pyramid with the base side \(a\) and the altitude \(H.\)

Solution.

Regular square pyramid with base side a and height H.
Figure 5.

The area of the square cross section at a point \(x\) is written in the form

\[A\left( x \right) = {\left( {a \cdot \frac{x}{H}} \right)^2} = \frac{{{a^2}{x^2}}}{{{H^2}}}.\]

Hence, the volume of the pyramid is given by

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

See more problems on Page 2.

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