Hello, Ana
The key to these volume problems is to break them down into two parts: The first part is figuring out what shape will result from the rotation, and the second part is figuring out how to calculate the volume of the shape. So, let's organize our process here into those two parts.
Part 1: Rotation
Now, we've got a region formed by the intersection of three functions:
y = 9x2
x = 1
y = 0
Before thinking about rotation, let's clarify what this 2-D region looks like.
- The bottom of this shape is a straight line formed by y = 0 (which we can also think of as the X-axis).
- The right side of this shape is also a straight line formed by x = 1.
- The third and final side is a concave arc formed by y = 9x2. It connects to the other two functions at the points (0,0) and (1,9).
So, our shape altogether looks kind of like a tall, very skinny triangle, except that it has a curved hypotenuse.
Now, we need to think about what's going to happen to this shape when we rotate it around the X-axis. For some people, they can imagine rotations easily without having to think too much. For most people, however, you need to have some kind of process for understanding the rotation process. What helps me with rotations is to re-imagine our region not as one whole shape but as a collection of many tiny lines and then think about what would happen to those lines when they rotate.
Here's what I mean by that. We already figured out that the region we are rotating looks kind of like a triangle. What I would do is re-imagine that triangle shape as a collection of vertical lines that all start at the bottom of our shape (formed by y=0) and rise straight up to a point on the function y =9x2.
So, the X-axis is the bottom point for all of these imaginary lines. When we rotate our region then, it's like the bottom of all of these lines stay fixed to the X-axis, but their top points rotate around the X-axis until they travel a full 360o and get back to their starting spots.
Now, think about what happens when you rotate a line around a point for a full 360o. You get a circle, with a radius equal to the line that was rotated. So, when we rotated our region, made up of all of these imaginary lines, what we made was a collection of circles all stacked against each other. This stack of circles starts at the origin (where y=9x2 intersected y=0) and continues up to x=1. The important part here, however, is that the size of each circle is determined by the length of the line that rotated around to form it (i.e. its radius). And, we already figured out that the length of those lines was the distance from the X-axis up to the corresponding point on the function y=9x2. So, our circles grow in radius from 0 up to 9, and our stack of circles actually looks like a funnel.
Part 2: Calculating Volume
Now, you might have been able to figure out the shape of our rotation without going through the work of dividing the region into lines. But, having done so will actually make the process of calculating volume much easier.
Very generally, we can think of volume as the integral of the area of a 2-D shape into 3-D space. Well, we just defined the volume of our rotation in terms of a 2-D shape: circles. Now we just have to figure out how to integrate them.
The area of circle is: A = πr2. In our case, however, the radius of each circle changes with respect to position X. Remember, the radius of each circle was the distance from the X-axis up to the corresponding point on the function y=9x2. So, we can actually define the radius as the function y=9x2. And, that's what we'll use to do our integration.
So V(x) = ∫Adx = ∫(πr2)dx = ∫(π(9x2)2)dx
- Which we can simplify: ∫(π(9x2)2)dx = ∫(π(81x4))dx = 81π∫(x4)dx
- Then, this integration is just a simple power rule. 81π∫(x4)dx = (81π) * ((x5)/5) = (81π/5)*(x5)
- But, we also know that this shape was rotated between x = 0 and x = 1. So, we'll use those values as the bounds to calculate this integral.
- (81π/5)*((1)5) - (81π/5)*((0)5) = (81π/5)*(1) - 0 = (81π/5)
Thus, we've found the Volume of our rotated region: (81/5)π
Best,
Daniel