# Thread: Volume of a solid obtained by revolution

1. ## Volume of a solid obtained by revolution

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=9x^2, x=1, y=0, about the x-axis

I know that 9x^2 is a parabola but what exactly am I supposed to do here? Am I trying to find the volume of what is on the right had side of the graph if I drew a vertical line at x=1 (kind of looks like a butterfly wing)?

Before anyone says to read the book, I did. This is an online calculus class and the professor refuses to answer questions from the students, he claims it is not his job to answer questions, just present the information and grade the tests.

Any help would be greatly appreciated.

2. Originally Posted by operaphantom2003
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=9x^2, x=1, y=0, about the x-axis

I know that 9x^2 is a parabola but what exactly am I supposed to do here? Am I trying to find the volume of what is on the right had side of the graph if I drew a vertical line at x=1 (kind of looks like a butterfly wing)?

Before anyone says to read the book, I did. This is an online calculus class and the professor refuses to answer questions from the students, he claims it is not his job to answer questions, just present the information and grade the tests.

Any help would be greatly appreciated.
Really? Sounds like a BAD teacher. Teachers are suppose to answer questions. You can't just throw a book at someone and demand that they learn calculus by themselves. That's NOT teaching.

When you draw the parabola, draw the line $x=1$ and then shade the area under the curve from $x=1$ to the origin. I'm pretty sure that this is the boundary. Imagine rotating this boundary around the x-axis. This is a solid of revolution.

The general way to deal with this is difficult to explain here.

Do you understand theory behind this?

The formula: $V=\lim_{n->\infty}A(x_i^*)\Delta x=\int_a^bA(x)dx$ where $A(x)$ is the cross-sectional area as a function of $x$ is what you need. See if this tutorial helps you understand how to find the cross-sectional area, and the theory behind this:

YouTube - Solid of Revolution (part 1)

In your case the volume should look like this:

$V=\int_0^1\pi (9x^2)^2dx$

The radius of the solid of revolution should be $9x^2$, so the cross-sectional area is $A(x)=\pi (9x^2)^2$. If this is confusing to you, make sure you watch the video I linked. Seeing someone demonstrate this will help you.

3. ## I still don't get it

I don't know whether to try more or just fail. This is what I have and it is wrong:
pi∫ (9x^2)^2 dx from 0 to 1
pi ∫ 9x^4 dx from 0 to 1
pi * 9x^5/5

plug in 1 for x and I get 9pi/5

But that is wrong. Where am I going wrong on this? I went to the tutoring center and they said this was really easy and it must be my fault for not understanding it. His actual words were " This is so easy I was doing it in junior high." Thanks....

Any help as to where I am wrong would be greatly appreciated.