# Thread: volume of solid revolution around y=3

1. ## volume of solid revolution around y=3

Find the volume of the solid that results when the region enclosed by $\displaystyle y=\sqrt{x}$, y=0 and x=9 is revolved about the line y=3.

Here is the image: Now the problem is that I can't even imagine the revolution. What figure would come up when the revolution is done?

Should the volume be $\displaystyle \int_{0}^{9} \pi (3-\sqrt{x})^2$
or maybe $\displaystyle \int_{0}^{9} \pi (3^2 - (3-\sqrt{x})^2)$

This is so strange.

It seems like the graphic of $\displaystyle \sqrt{x}$ is below the y=9 axis, which is not actually consistent with the definition.

It is like revolution of negative function around the x-axis. Very strange.

2. Originally Posted by p0oint Find the volume of the solid that results when the region enclosed by $\displaystyle y=\sqrt{x}$, y=0 and x=9 is revolved about the line y=3.

Here is the image: Now the problem is that I can't even imagine the revolution. What figure would come up when the revolution is done?

Should the volume be $\displaystyle \int_{0}^{9} \pi (3-\sqrt{x})^2dx$

That's the volume of the internal conical part, because the radius of the discs is the difference between y=3 and the function.
This will calculate the "conical" volume first, which can be subtracted from a cylinder.
An indirect way to calculate the volume you want.

or maybe $\displaystyle \int_{0}^{9} \pi (3^2 - (3-\sqrt{x})^2)$

This is so strange.

It seems like the graphic of $\displaystyle \sqrt{x}$ is below the y=9 axis, which is not actually consistent with the definition.

It is like revolution of negative function around the x-axis. Very strange.
The "internal part is a "cone" shape, with the "opening" against the y-axis from y=0 to y=6.

y=3 is the axis of symmetry, hence imagine it made from a series of vertical discs
(you know what a tornado looks like!).
The radius of each disc is the distance from the function to the line y=3.
Hence the volume of the conical part is the integral from x=0 to x=9 of the disc areas.

Then subtract that answer from the volume of the external cylinder of radius 3 and height 9.

Think of the graph of the function from x=0 to x=9 as a line which is then made spin around the axis y=3.

3. One way to think about it is

$\displaystyle \pi \int_a^b r_o^2 - r_i^2 dx$
where $\displaystyle r_o$ is the outer radius of a disc from the axis of symmetry (in your caase $\displaystyle r_o = 3-\sqrt{x}$)
and $\displaystyle r_i$ is the inner radius (if you has a second curve).

4. I understand what you are saying to me. But look at the region.

The revolution is not cone. It is like cylinder with cone mould out of the cylinder.

So we need to subtract the volume of cylinder out of the volume of cone.

Am I right?

5. Originally Posted by p0oint I understand what you are saying to me. But look at the region.

The revolution is not cone. It is like cylinder with cone mould out of the cylinder.

So we need to subtract the volume of cylinder out of the volume of cone.

Am I right?
No, the revolution of the volume you are looking for is not a cone.
But the region within it is (not a uniform cone of course!).
Both regions are within a cylinder of diameter 6 and height 9.
Hence you can alternatively calculate the volume of revolution of the conical part
and subtract that from the cylinder.

The shape in question is what is left after removing the "cone".

Danny showed how to find the required volume directly,
which you were also working out the integral for.

6. Thank you.

Yes, I know Danny what is saying, so I posted in the first post:

$\displaystyle \int_{0}^{9} \pi (3^2 - (3-\sqrt{x})^2)$

This is correct, right?

7. Originally Posted by p0oint Thank you.

Yes, I know Danny what is saying, so I posted in the first post:

$\displaystyle \int_{0}^{9} \pi (3^2 - (3-\sqrt{x})^2)dx$

Yes, don't forget your "dx" notation.

This is correct, right?
Exactly, the cross-sectional area of the shape is the difference between 2 discs.

8. Originally Posted by p0oint Thank you.

Yes, I know Danny what is saying, so I posted in the first post:

$\displaystyle \int_{0}^{9} \pi (3^2 - (3-\sqrt{x})^2)$

This is correct, right?
Here's a picture to go with it (sorry for the quality). Here $\displaystyle r_0 = 3$ and $\displaystyle r_i = 3 - \sqrt{x}$ so yes, you are correct.

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# find the volium of the solid that results when the region enclosed by y=√x ,y=0 and x=9 is revolved about the line x=9

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