# 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 $$A\left( x \right)$$ for $$a \le x \le b.$$

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

Click or tap a problem to see the solution.

### 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.

### 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 $$z = 1.$$

### 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.

### Example 4

Find the volume of a regular square pyramid with the base side $$a$$ and the altitude $$H.$$

### 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.

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.

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.

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.

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.