Computing Volumes of Surfaces of Revolution

Everybody knows that the voluem of a solid box is

   volume$\displaystyle =$   length$\displaystyle \times$   width$\displaystyle \times$   height$\displaystyle .

More generally, the volume of cylinder is $ V = \pi r^2 h$ (cross sectional area times height). Even more generally, if the base of a prism has area $ A$, the volume of the prism is $ V=A h$.

But what if our solid object looks like a complicated blob? How would we compute the volume? We'll do something that by now should seem familiar, which is to chop the object into small pieces and take the limit of approximations.

[[Picture of solid sliced vertically into a bunch of vertical thin solid discs.]]

Assume that we have a function

$\displaystyle A(x) =$   cross sectional area at $x$$\displaystyle .

The volume of our potentially complicated blob is approximately $ \sum A(x_i) \Delta x$. Thus

volume of blob $\displaystyle = \lim_{n\to\infty} \sum_{i=1}^n A(x_i) \Delta x$    
  $\displaystyle = \int_{a}^b A(x) dx$    

Example 3.2.1   Find the volume of the pyramid with height $ H$ and square base with sides of length $ L$.

Figure 3.2.1: How Big is Pharaoh's Place?
Image pyramid

For convenience look at pyramid on its side, with the tip of the pyramid at the origin. We need to figure out the cross sectional area as a function of $ x$, for $ 0\leq x \leq H$. The function that gives the distance $ s(x)$ from the $ x$ axis to the edge is a line, with $ s(0)=0$ and $ s(H) = L/2$. The equation of this line is thus $ s(x) =
\frac{L}{2H} x$. Thus the cross sectional area is

$\displaystyle A(x) = (2s(x))^2 = \frac{x^2L^2}{H^2}.

The volume is then

$\displaystyle \int_{0}^{H} A(x)dx =
\int_{0}^{H} \frac{x^2L^2}{H^2} dx
= \left[ \frac{x^3L^2}{3H^2}\right]_{0}^H
= \frac{H^3L^2}{3H^2} = \frac{1}{3} HL^2.

Today: Quiz!
Next: Polar coordinates, etc.
Recall: Find volume by integrating cross section of area. (draw picture)

Example 3.2.2   Find the volume of the solid obtained by rotating the following region about the $ x$ axis: the region enclosed by $ y=x^2$ and $ y=x^3$ between $ x=0$ and $ x=1$.
Figure 3.2.2: Find the volume of the flower pot
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The cross section is a ``washer'', and the area as a function of $ x$ is

$\displaystyle A(x) = \pi(r_o(x)^2 - r_i(x)^2) = \pi (x^4 - x^6).

The volume is thus

$\displaystyle \int_{0}^1 A(x) dx = \int_{0}^1 \left(\frac{1}{5}x^5 - \frac{1}{7...
...= \pi \left[\frac{1}{5}x^5 - \frac{1}{7}x^7\right]_{0}^{1}
= \frac{2}{35}\pi.

Example 3.2.3   One of the most important examples of a volume is the volume $ V$ of a sphere of radius $ r$. Let's find it! We'll just compute the volume of a half and multiply by $ 2$.
Figure 3.2.3: Cross section of a half of sphere with radius 1
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The cross sectional area is

$\displaystyle A(x) = \pi r(x)^2 = \pi (\sqrt{r^2-x^2})^2 = \pi (r^2-x^2).


$\displaystyle \frac{1}{2} V = \int_{0}^r \pi (r^2-x^2) dx
= \pi \left[ r^2 x - \frac{1}{3}x^3\right]_0^r
= \pi r^3 - \frac{1}{3} \pi r^3 = \frac{2}{3}\pi r^3.

Thus $ V = (4/3) \pi r^3$.

Example 3.2.4   Find volume of intersection of two spheres of radius $ r$, where the center of each sphere lies on the edge of the other sphere.

From the picture we see that the answer is

$\displaystyle 2 \int_{r/2}^{r} A(x),

where $ A(x)$ is exactly as in Example 3.2.3. We have

$\displaystyle 2 \int_{r/2}^{r} \pi (r^2-x^2) dx = \frac{5}{12} \pi r^3.

William Stein 2006-03-15