In fact, visualizing some of those objects in your head will probably be the trickiest part of integrating by slices. The examples below will help to get you used to thinking about 3-D objects with simple cross sections.

The idea

---

The basic idea is that if an object can be sliced into an infinite number of infinitesimally thin slices (thinner than these apple slices), and if those slices have a predictable 2-D shape (near circles for the apple), we can add those up by integration to get the volume.

If you wanted to find the volume of a shoebox, for example, you could add up rectangular slices along any one of three different directions, x, y or z.

In the examples that follow, notice that we're working in 3-D (x, y, z) coordinates.

The way we'll determine the volume of this object is to note that cross sections perpendicular to the x-axis, like the one in blue, have area = h x w, and infinitesimal width of dx. If we then use the integral along the x-axis as the sum of an infinite number of these rectangles, we get the volume of the box:

$$V = \int_0^l h\cdot w \, dx = hw\cdot x \bigg|_0^l = hwl$$

which matches our expectation of the volume. We could also have found the same result by summing up rectangles in the x-y plane along the z-axis or by summing rectangles in the x-z plane along the y-axis. You might want to set up those integrals for yourself and make sure you get the same result: V = h·w·l.

We have all of the information we need to calculate the area of one of the squares as a function of x or y. We start by finding the length of a side of the square as a function of x. We use x because it's the direction along which we'll be stacking squares to fill the volume:

$$x = y^2 \; \longrightarrow \; y = \sqrt{x}$$

Now the side of the square is twice that (see the blue notation in the diagram):

With the area in hand, we now just integrate along the x-axis from the origin (x = 0) to x = 2:

$$ \begin{align} V &= \int_0^2 (2 \sqrt{x})^2 \, dx \\ \\ &= \int_0^2 \, 4x \, dx = 2x^2 \bigg|_0^2 = 8 \; units^3 \end{align}$$

Notice that x was the obvious direction along which to integrate in this example. The cross sections of this object perpendicular to the x-axis can be predictably described using the parabola equation.

Then by simple algebra, we have:

$$h = \sqrt{3x^2} = x\sqrt{3}$$

Now we're integrating along the y-axis, so we'll want to express h and x (and ultimately the area of each triangle) as a function of y. We begin by rearranging the circle equation:

$$ \begin{align} x^2 + y^2 &= r^2 \\ x &= ±\sqrt{r^2 - y^2} \end{align}$$

Then the area of our equilateral triangle is twice the area of one of the right triangles:

After we insert x as a function of y, we get

$$A = (r^2 - y^2) \sqrt{3}$$

which we can integrate. Notice that we can simplify the integral by taking advantage of the symmetry of a circle. It's always easier to evaluate a definite integral when one of the limits is zero:

This is a pretty easy integral. Here's how to finish it and evaluate the limits:

$$ \begin{align} &= 2 \sqrt{3} \left(r^2y - \frac{y^3}{3} \right) \bigg|_0^r \\ \\ &= 2 \sqrt{3} \left(r^3 - \frac{r^3}{3} \right) \\ \\ &= 2 \sqrt{3} \left(\frac{2r^3}{3} \right) = \frac{4r^3\sqrt{3}}{3} \end{align}$$

In problems like this, when no units are specified, it's a good idea to be in the habit of writing "units³" in the place of actual units. That way, when there are units, you'll be less likely to forget them.