Hi there again, I have not written a post in while, here goes my second post.

I would like us to discuss one of the important applications of integration. We have seen how integration can be used to solve the area problem, in this post we are going to see how we can use a similar idea to solve the volume problem. I suggest that we start by looking at the solids whose volume we know very well. You should be able to calculate the volumes of the cylinders below (yes, they are all cylinders.)

Cylinders are nice, we only need to multiply the cross-sectional area by the height/length to find the volume. This is because they have two identical flat ends and the same cross-section from one end to the other. Unfortunately, not all the solid figures that we come across everyday are cylinders. The figures below are not cylinders.

How do we calculate the volume of a non-cylindrical solid figure (how do we calculate the area of a non-rectangular object?)? We chop the solid figure into infinitely many cylinders (we chop the object into infinitely many rectangles). Summer is coming, we would like to know how much ice cream those cones can hold! Let’s go right ahead and chop the cone into infinitely many cylinders. We will put the cone with base radius and length(height) in the Cartesian plane like below:

We start by chopping off a typical circular cylinder at some arbitrary value, . We then sum up all the volumes of the cylinders across the entire cone, the thinner the cylinders the better. The volume of an arbitrary cylinder with thickness (or height) , is given by

but the radius is actually the -value at that arbitrary value, so the volume is

Let’s sum up all these n cylinders (like we summed up all the rectangles when we were introducing the integral) to get the total volume.

If we take infinitely many cylinders , we get the integral,

Yay! Now we are down to integration, our food. We do realize that we need to write the in terms of because we are integrating with respect to . Back to the Cartesian plane, how is related to ? We have a straight line going through the origin, and passing through a point , we should be able to show that the equation of this straight line is this makes , . The values are running from to , the volumes becomes:

After enjoying our food (ice cream), we get that the volume of a cone with height and base radius is

.

We should be able to use the same technique to derive the formula for the volume of a sphere by placing the center of the sphere at the origin. Please do it!

This was nice, I’ll be writing another post soon!

UCT MAM1000W lecture notes subject links – MathemafricaJanuary 17, 2018 at 10:07 am[…] More volumes of integration […]