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5.1 Volumes of solids

Recall from Section 4.1 that the double integral

I=\int\!\int_{R}f(x,y)\;dx\;dy

is the volume between the region $R$ and the surface $z=f(x,y)$ (with any volume below the $(x,y)$ -plane counted negatively).

The following result can often be used to find the volume of simple bodies and their rearrangements. A good example of a body is a loaf of bread; often this is cut into slices which are parallel.

Cavalieri’s Slicing Principle. Let $B$ be a body and $P_{z}$ a family of parallel planes such that:

(i)

for $u\leq z\leq w,$ the plane $P_{z}$ lies between $P_{u}$ and $P_{w}$ ;
(ii)

$B$ lies between $P_{a}$ and $P_{b}$ ;
(iii)

the area of the slice of $B$ that is cut by $P_{z}$ is $A(z)$ .

Then the volume of $B$ is $V(B)=\int_{a}^{b}A(z)\,dz.$

Thus the volume of a body is the integral of the areas of its parallel slices. We can also think of this as a result about rearranging bodies so that the volume becomes easier to calculate. If we have two loaves of bread such that the slices have the same area and thickness, then the loaves contain an equal amount of bread.

Corollary. Suppose that two bodies can be positioned upon a plane $P$ such that each plane parallel to $P$ intersects both bodies in slices that have equal areas. Then the volumes of the two bodies are equal.

Example. Find the volume enclosed in the tetrahedron that has vertices $(0,0,0),(1,0,0),(0,1,0)$ and $(0,0,1).$

The cross-section of the pyramid at height $z$ is an isosceles triangle with two sides (at right angles) of length $(1-z)$ . Thus its area is $\frac{1}{2}(1-z)^{2}$ . By 5.18, the volume of the pyramid is

$\displaystyle{\int_{0}^{1}\frac{1}{2}(1-z)^{2}\,dz=}$

Example. Show (by slicing) that the volume enclosed by a sphere of radius $r$ is $4\pi r^{3}/3.$

The equation of the sphere is $x^{2}+y^{2}+z^{2}=r^{2}$ . Thus the cross-section of the sphere at a fixed $z\in(-r,r)$ is given by $x^{2}+y^{2}=r^{2}-z^{2}$ , which is a circle of radius $\sqrt{r^{2}-z^{2}}$ . The area of this cross-section is therefore $\pi(r^{2}-z^{2}).$

By Cavalieri’s Slicing Principle, the volume of the sphere is

\pi\int_{-r}^{r}(r^{2}-z^{2})\,dz=\pi\left[r^{2}z-\frac{z^{3}}{3}\right]_{-r}^% {r}=\,2\pi\left[r^{2}z-\frac{z^{3}}{3}\right]_{0}^{r}=

Solids of revolution.

A solid of revolution is a body generated by rotating a two-dimensional curve about an axis. Examples include spheres, cylinders, cones and paraboloids. Cavalieri’s Slicing Principle provides a simple formula for computing the volume of a solid of revolution.

Suppose $f$ is a function of $x$ . Then the volume of the solid obtained by rotating the portion of the graph between $x=a$ and $x=b$ about the $x$ -axis is given by

V=\int_{a}^{b}\pi(f(x))^{2}\,dx

because any cross-section of the solid taken perpendicular to the $x$ -axis is a circle of radius $f(x)$ whose area is equal to $\pi(f(x))^{2}$ .

Example. Compute the volume enclosed by a cone of length $h$ and radius $r$ .

The gradient of the straight line is $\tan\theta=r/h$ , and hence the equation of the line is $f(x)=rx/h$ . The limits of integration are from $x=0$ to $x=h$ . Therefore, the volume of the cone is

V=\int_{0}^{h}\pi\left(\frac{rx}{h}\right)^{2}\,dx=\pi\left[\frac{r^{2}x^{3}}{% 3h^{2}}\right]_{0}^{h}=\frac{\pi r^{2}h}{3}.

Volumes enclosed by intersecting surfaces.

Let $R$ be a region in the $x y$ plane and take two surfaces $z_{1}=z_{1}(x,y)$ and $z_{2}=z_{2}(x,y)$ , and suppose that they intersect in such a way that the points between the surfaces form a body

B=\{(x,y,z):z_{1}(x,y)\leq z\leq z_{2}(x,y);(x,y)\in R\}

which lies above the region $R$ in the plane. Then the volume of this body is

{\hbox{vol }}(B)=\int\!\!\!\int_{R}(z_{2}-z_{1})\,dxdy.

Example. Compute (again) the volume inside the sphere $x^{2}+y^{2}+z^{2}=1$ .

We consider the two halves of the sphere $z_{1}=-\sqrt{1-x^{2}-y^{2}}$ and $z_{2}=\sqrt{1-x^{2}-y^{2}}$ . Then the sphere is the set of points $(x,y,z)$ with $x^{2}+y^{2}\leq 1$ and $z_{1}(x,y)\leq z\leq z_{2}(x,y)$ . Therefore, the volume of the sphere is

\int\!\!\!\int_{D}(z_{2}-z_{1})\,dxdy=2\int\!\!\!\int_{D}(1-x^{2}-y^{2})^{1/2}% \,dxdy

where $D=\{(x,y):x^{2}+y^{2}\leq 1\}$ is the disc in the plane of unit radius and centre $(0,0)$ . Thus we can write the double integral as

2\int_{-1}^{1}\int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}}(1-x^{2}-y^{2})^{1/2}dxdy.

To calculate this integral, we make the substitution $x=\sqrt{1-y^{2}}\sin t$ , so that $\frac{dx}{dt}=\sqrt{1-y^{2}}\cos t$ and

$\displaystyle{1-x^{2}-y^{2}=}$

The bounds $-\sqrt{1-y^{2}}\leq x\leq\sqrt{1-y^{2}}$ are given by $\qquad\qquad\qquad$ . Thus the inner integral is

$\displaystyle{\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(1-y^{2})\cos^{2}t\,dt=}$

Thus the volume is

$\displaystyle{\pi\int_{-1}^{1}(1-y^{2})\,dy=}$