Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
x=6y^2, y=1, x=0 about the y-axis

Answer 1

Well first, you need to have an idea what this giddy-up look like. In x=6y^2 you can solve for y and punch in it your calculator

Answer 2

I've been trying for a while now, and I can't figure it out

Answer 3

What did you get?

Answer 4

I got 3. I'm not sure how to set it up.

Answer 5

Well you should still have an x.
x=6y^2
x/6=y^2
(sq rt x)/(sq rt 6)=y

Answer 6

I shouldn't get an x in the final answer... My final answer was 3.

Answer 7

\[\int\limits_{0}^{1}Pix ^{2}dy\]

Answer 8

You are just manipulating the equation, so that it fits into your calculator at the y= button. You don't get rid of x unless you are solving for a specific something and then you would have a value of x to enter. This is just the preliminary stage to find out what the graph looks like.

Answer 9

I can't even get to that point. I'm so lost.

Answer 10

Don't worry about it. I already gave you. Put in your calculator graphing function y=(square root of x)/(square root of 6)

Answer 11

Okay I got the graph.... What do I do with that?

Answer 12

Draw that line on a piece of paper and then draw the line y=1. The graph they are describing is the space between the thing from your calculator, x=0, y=1

Answer 13

I got that far before and got the answer 3.

Answer 14

Well, I don't know what the answer is. I was just trying to get to visualize what the problem is saying, which is the first step in solving these kind of problems.

Answer 15

I can visualize it; I just don't know how to set it up. Thank you though.

Answer 16

I am showing you how to do it and any problem like it.

Answer 17

Think of it as making disks. The volume of a cylinder is pi*r^2*h. If you have your graph as f(y) and are rotating around the y axis, then imagine: cylinder with height dy and radius f(y). So you sum up these cylinders by integrating from y=0 to y=1 using the formula for the volume of a cylinder. That's how im visualizing it.
So i'm getting what duct said