# Calculus

## Applications of Integrals # 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 $$z = \large{\frac{{{x^2}}}{{{a^2}}}}\normalsize + \large{\frac{{{y^2}}}{{{b^2}}}}\normalsize$$ 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 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.

### 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 $$H.$$

### Example 7

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

### Example 8

Find the volume of a regular tetrahedron with the edge $$a.$$

### 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.

### 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.

### 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 $$z = \large{\frac{{{x^2}}}{{{a^2}}}}\normalsize + \large{\frac{{{y^2}}}{{{b^2}}}}\normalsize$$ 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:

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

### 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

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

### 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.

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.

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:

${\text{major axis:}\;\;a + \left( {A – a} \right)\frac{x}{H}\;\;}\kern0pt{\text{minor axis:}\;\;b + \left( {B – b} \right)\frac{x}{H}}.$

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

${A\left( x \right) \text{ = }}\kern0pt{\pi \left( {a + \left( {A – a} \right)\frac{x}{H}} \right)\left( {b + \left( {B – b} \right)\frac{x}{H}} \right) }={ \pi \left[ {ab + b\left( {A – a} \right)\frac{x}{H} }\right.}+{\left.{ a\left( {B – b} \right)\frac{x}{H} }\right.}+{\left.{ \left( {A – a} \right)\left( {B – b} \right){{\left( {\frac{x}{H}} \right)}^2}} \right] }={ \pi \left[ {ab + \left( {bA + aB – 2ab} \right)\frac{x}{H} }\right.}+{\left.{ \left( {AB – aB – bA + ab} \right){{\left( {\frac{x}{H}} \right)}^2}} \right].}$

Then the volume is given by

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

### Example 7.

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

Solution.

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

${m = c + \left( {a – c} \right)\frac{x}{h},\;\;}\kern0pt{n = \frac{{bx}}{h}.}$

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.

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

$\require{cancel}{m = \frac{{ax}}{h} }={ \frac{{\sqrt 3 \cancel{a}x}}{{\sqrt 2 \cancel{a}}} }={ \frac{{\sqrt 3 x}}{{\sqrt 2 }}.}$

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 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 $$\large{\frac{1}{8}}\normalsize$$ of the solid of intersection.

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

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