Volume of a Solid of Known Cross Section
Updated: Jan 9, 2020
Find the volume of the solid with these parameters.
Since we have been given both boundaries for a base and the general shape of the cross sections, we will be using the method of: “volume of a solid with known cross section”. The best way to tackle these types of problems, or at least the way that has helped me the most, is to first draw what the problem is talking about.
The base is the parabola y = x^2 cut off by y = 3, and the cross sections are squares that are perpendicular to the y -axis (so the side of the square touching the y-axis is perpendicular to the y-axis).
NOTE: Whenever you draw a 3D coordinate axis, make sure it follows the right hand rule: so if you were to stick out your right hand so that it is parallel to the x-axis and curl your hand towards the y-axis, your thumb should point in the direction of the z-axis.
Once we have that down we can start solving the problem. To find the volume with this technique, we basically slice up the solid into a bunch of those cross sections (so in this case a bunch of squares), and then add them all up.
To be as accurate as possible, we want the thickness of each of these cross sections to be infinitesimally small/as thin as possible, so when we add them all up we get the actual volume. Whenever you are adding up a bunch of very small segments or values over an interval, you have to use an integral. So the volume of known cross section is just:
So which variable, x or y, do we use to integrate?
Generally speaking, you use the variable that the cross sections are perpendicular to, so we would use y.
You could also see this from the drawing. This is because if the cross section is perpendicular to the y-axis, then the infinitesimally small width/thickness of your cross sections are parallel to the y-axis (so it’s a small width in the y-direction)
So now we know that we have:
Now we need to find what A(y) is and what the interval [a,b] is.
We know the interval should be in terms of y since we are integrating with respect to y. Visually, this means we have to see between which two points the solid goes along the y axis. We know the base is bounded by y = x^2, which starts at y = 0 (since if you were to just graph a parabola it would start at the origin), and bounded by y = 3. So the interval is just [0,3].
For area, we know that the area of a square is s^2, where s is the length of one side. And we can find s by looking at our drawing.
We can see that the length of one side goes from one end of the parabola to the other parallel to the x-axis. And since parabolas are symmetric, that means it goes from -x to x, which means:
s = 2x.
But out integral is in terms of y, so we need to have s in terms of y as well. To do that, we use the fact that y = x^2. We can solve for x and get x = sqrt(y). So s = 2*sqrt(y). We use this and not y = 3 since the sides of the square are bounded by the parabola.
Now we have all that we need to solve the problem: