FFR - Volumes

Application of Integrals

Integration connects rates with changes. If s(t) is position and v(t) is velocity, then ∫_a^b v(t) dt = s(b) − s(a), the displacement. Total distance travelled is ∫_a^b |v(t)| dt; split at velocity sign changes when needed. Similarly, integrating acceleration gives a change in velocity, and an initial velocity is needed to recover the velocity function.

In addition to physics, integrals also have many other applications. So far we have been taking single integrals, which give us areas. Double integrals are used in Calculus 3 to give complex volumes.

This is not a calculus 3 book, but we can still use single integrals to get simpler volumes. What kind of volumes? Mostly prisms and rotated shapes. All prism volumes are derived from integrals! The same goes for the volume of a sphere and other symmetric spherical shapes.

For example, take the volume of a very familiar shape–a cylinder. Its volume is simply the area of its circle shape on top, multiplied by its height (in other words, the area of the circle is “integrated” from 0 to h).

There is a special name for the circular shape on top of the cylinder. It’s called a disc because it is infinitely thin and in the shape of a circle. Whereas before we were integrating really thin rectangles, now that we are finding volumes, we work with discs.

As a point of comparison, recall that the formula for the volume of a cylinder is: $V=\pi r^{2}h$.

The integral for solving this volume problem would look something like this:

${\int }_1^6\pi \cdot f(x){}^{2}dx$

$f(x)$ represents the “radius,” but the equation holds true whether $f(x)$ is a constant function (like this case) or whether it is any other function. If we were trying to find the volume of a cone, $f(x)$ might be $y=-0.5x+2$.

$dx$ represents something different this time. Whereas before it represented really tiny, infinitely thin rectangles (you can almost think of them as line segments, that’s how thin they are), now it represents infinitely thin rotated discs. Integrating these discs over an interval gives us the entire volume.

Finally, the interval of integration is equal to the height. Using this general integral, we will be able to derive the volume of cones and even spheres.

An alternate method of finding volumes

The previous method we discussed is also known as the method of discs, because we sum up the area of each individual disc to find the total volume. The next method is known as the method of cylindrical shells, because we divide up the cylinder into smaller “shells” and find the volume by summing the area of these infinitely thin shells. It looks something like this:

$V={\int }_0^{3}x\cdot 2\pi ydy$

Now, $2\pi ydy$ is equal to the summation of the circumference of the cylindrical shells. Notice that the circumference of one shell is different from the circumference of the next shell.

$x$ now represents the height. In this case, it is constant (6-1=5) for all shells, but for more complex shapes like cones, this may have to be rewritten as a function of y so that the integral is defined.

Also notice that the intervals of integration have changed. Some of you might wonder why it’s not from -3 to 3, and it’s because the intervals of integration represent the radius (“y”). The outermost shell has a radius of 3, not 6.

Which method is better? Each has its advantages. The rule of thumb is that if a shape is rotated around the y-axis, cylindrical shells is usually better. Conversely, if a shape is rotated around the x-axis, the method of discs is usually a better choice because integrating in the y-direction is not something most of us are familiar with.

Imagine how the formula for the cylindrical shells method might be different if the cylinder was centered around the y-axis rather than the x-axis, with the same radius and height? It would look something like this:

$V={\int }_0^{3}y\cdot 2\pi xdx$​
This time, y is the constant height, and x is the varying radius. Again, for more complex functions, y represents a function of x, so the integral would naturally be more complicated for shapes like cones or parabolic cylinders.

Let’s solve both of the integrals we discussed earlier to see whether they both give us the same volume.

Discs:

$V={\int }_1^6\pi \cdot f(x){}^{2}dx$​
In this case, f(x) is a constant function ($y=3$), so we integrate $9\pi$ with respect to x from 1 to 6.

${\left[9\pi x\right]}_1^6$

$9\pi (6-1)$

​
$=\mathbf{45}\pi$​

Cylindrical Shells:

$V={\int }_0^{3}x\cdot 2\pi ydy$

Recall that x is constant and equal to 5 here, so we can factor it out of the integral. Evaluating the integral gives us

$=\cancel{10}{}^5\pi \cdot {\left[\frac{y^2}{2}\right]}_0^3$

$=5\pi \cdot (9-0)$

$=\mathbf{45}\pi$

Both methods give us a consistent answer!

Chapter 9 Problems

1. Solve the indefinite integral $\int \csc xdx$
2. Evaluate ${\int }_0^{\pi }\frac{\cos x}{\cos (\sin x)}dx$

3. Evaluate ${\int }_0^{2}6\sqrt{4-\sqrt{x}}dx$
4. Solve the indefinite integral $\int \frac{2}{3}\sqrt{9-x^2}dx$

5. Evaluate the indefinite integral $\int \frac{1-4x^2}{x}dx$
6. Evaluate ${\int }_0^{\mathrm{\infty }}\frac{2-x}{8-4x+2x^2-x^3}dx$

   Clarification: Interpret the removable hole at x=2 by taking the finite one-sided limits, or equivalently by using the continuous extension.

7. Evaluate ${\int }_1^3\frac{x\cdot \sqrt{2-x+\frac{3}{x}}}{\sqrt{x}}dx$
8. Evaluate the indefinite integral $\int \tan x\cdot \frac{\sin (\ln (\cos x))}{\cos (\ln (\cos x))+3}dx$

9. Evaluate the indefinite integral $\int \frac{\tan (\ln x)}{x\cdot \ln (\cos (\ln x))+3x}dx$
10. Solve $\int 1-2x\sqrt{1-x^2}dx$