Volume by rotating the y axis

**matt3D** · September 24th 2008, 09:35 PM

Hi, here is my problem: Find the volume formed by rotating the region enclosed by:

and

with

about the

-axis.
So I get:
$V=\ 2\pi\int_0^{\sqrt6} (y^3)^2-(6y)^2 dy$
But my answer isn't right. Is this the correct way to set it up?
Thanks,
Matt

**Chris L T521** · September 24th 2008, 10:04 PM

Originally Posted by matt3D

Hi, here is my problem: Find the volume formed by rotating the region enclosed by:

and

with

about the

-axis.
So I get:
$V=\ 2\pi\int_0^{\sqrt6} (y^3)^2-(6y)^2 dy$
But my answer isn't right. Is this the correct way to set it up?
Thanks,
Matt

It should be $\pi\int_0^{\sqrt{6}}\left[(6y)^2-(y^3)^2\right]\,dy$

Can you see why this is the proper set up?

--Chris

**matt3D** · September 24th 2008, 10:20 PM

Oh okay, I think I see. This is the disk method right? So it's the top function minus the bottom right...? I think I was getting confused with the shell method... hehe
Here is a picture of the graph I did in mathematica:

**Chris L T521** · September 24th 2008, 10:26 PM

Originally Posted by matt3D

Oh okay, I think I see. This is the disk method right? So it's the top function minus the bottom right...? I think I was getting confused with the shell method... hehe
Here is a picture of the graph I did in mathematica:

You have the right idea! Keep in mind, though, that the horizontal refers to values of y, not x. So you would take the top curve $x=6y$ , square it, and subtract it from the square of the lower curve $x=y^3$

Is it clearer now on why the volume [using the washer method] would be $\pi\int_0^{\sqrt{6}}\left[(6y)^2-(y^3)^2\right]\,dy$ ?

--Chris

**matt3D** · September 24th 2008, 10:42 PM

Okay, I think I understand where you're coming from. What do you mean exactly when you said, "that the horizontal refers to values of y, not x." So when I have a problem that says rotate about the y axis, the limits are referring to y correct? But what about something like this:
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

about the y-axis

So would I use the washer method and integrate from 0 to 1?

**Chris L T521** · September 24th 2008, 10:56 PM

Originally Posted by matt3D

Okay, I think I understand where you're coming from. What do you mean exactly when you said, "that the horizontal refers to values of y, not x."

According to your graph, the horizontal values were the y values. Otherwise it would look like this if you were looking at it from the standard cartesian system:

So when I have a problem that says rotate about the y axis, the limits are referring to y correct? But what about something like this:
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

about the y-axis

So would I use the washer method and integrate from 0 to 1?

Get these in terms of y:

$y=x^2\implies x=\sqrt{y}$

So, in terms of y, $R(y)=3$ and $r(y)=\sqrt{y}$ , and its being integrated from 0 to 9. [Since at x=3, y=9]

So the integral here would be $\pi\int_0^9\left[(3)^2-(\sqrt{y})^2\right]\,dy=\pi\int_0^9\left[9-y\right]\,dy$

Does this make sense?

--Chris

**matt3D** · September 24th 2008, 11:59 PM

Yea, I think so, so you subtract $3-\sqrt{y}$ because you do the right function minus the left and square both. But how will it look on a graph, what portion gets thrown around. I couldn't figure out how to plot a straight line for x=3 so I just drew it in photoshop.

**Chris L T521** · September 25th 2008, 09:35 AM

Originally Posted by matt3D

Yea, I think so, so you subtract $3-\sqrt{y}$ because you do the right function minus the left and square both. But how will it look on a graph, what portion gets thrown around. I couldn't figure out how to plot a straight line for x=3 so I just drew it in photoshop.

The light green shaded region is what were are dealing with here:

Since we graphed the functions as $x=f(y)$ , the horizontal here represents the y axis and the vertical represents the x axis.

So here, we have the top function as $x=3$ , and the bottom function is $x=\sqrt{y}$ . These two functions intersect at $y=9$

Since we are revolving it about the y axis [horizontal], our volume integral would be $\pi\int_0^9\left[(3)^2-(\sqrt{y})^2\right]\,dy=\pi\int_0^9\left(9-y\right)\,dy$

Does this make sense?

--Chris

**matt3D** · September 25th 2008, 12:03 PM

OHhhhhhhhhhhhh, thank you! I forgot that you could change the axises around.

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