# Thread: Volumes of solid revolutions

1. ## Volumes of solid revolutions

I'm fed up with solid revolutions and shapes. I've tried examples and it's not helping with the particular problem I'm trying to solve.

Here it is:
Use the washer method to compute the volume of the solid obtained by rotating the region bounded by the curves y = 1-X^(2), y = 0

This is what I came up with, inner radius= 1-0x, outer radius = 1-(1-x^(2)) or x^(2) when calculated out.

so... A(x) =pi (x^(2)+0)^(2)-(1-0x)^(2)) ==== pi integrated on bounds of -1 to 1(x^(4)-1)dx ===== turns into (1/5^(5)-1x) when I take the anti derivative. I put in 1 and -1 doing fundamental theorem of calc and end up with 5.026548.

Is this even close?

There is another problem that is on my sheet. The only variation is instead of y=0, it has y=-1 instead.

If anyone could help me with solids of revolution it would be GREATLY appreciated. I passed calc I with a B, but I am having quite a lot of trouble right now with this class, I am trying my butt of and no matter what it feels like my efforts aren't good enough. I almost don't know what to do......

2. ## Re: I'm just incredibly frustrated right now with volumes of solid revolutions

Originally Posted by tastybrownies
Here it is:
Use the washer method to compute the volume of the solid obtained by rotating the region bounded by the curves y = 1-X^(2), y = 0

A (about the x axis) ...
rotating the region between $y = 1-x^2$ and $y = 0$ about the x-axis would involve disks, not washers ... either you 're using the wrong technique, or something is missing from your problem statement.

$V = \pi \int_{-1}^1 (1-x^2)^2 \, dx$

rotating the same region about $y = -1$ would involve washers ... note the cylindrical "hole" formed by the rotation.

$V = \pi \int_{-1}^1 [(1-x^2) - (-1)]^2 - 1^2 \, dx$

3. ## Re: I'm just incredibly frustrated right now with volumes of solid revolutions

Yeah, not sure why it says to use the washer method on the worksheet, maybe a mistake?

Well, took your advice and used the other method, A(x) = pi x^(2) change in x for the disks.

(-x^(2)+1)^(2) == x^(4)-2x^(2)+1, after integrating I ended up with

pi * (1/5(x)^(5)-2/3x^(3) )+1x/ integrated from -1 to 1,

I ended up with (8/5)- (-8/5) *pi = 16pi/15 or 3.35....

For the other method, I did
(x^(2)+1)^(2) - (2-0x)^(2)
after multiplying and integrating this, i ended up with the equation,
1/5x^(5) - 2/3x^(3) -3x/ integrated from -1 to 1,

I ended up with -6.93*pi= -21.78. Not sure if this should automatically be an absolute value because its volume.

Thank you again, I'm just trying to understand and wrap my mind around this!

4. ## Re: I'm just incredibly frustrated right now with volumes of solid revolutions

Originally Posted by tastybrownies
Yeah, not sure why it says to use the washer method on the worksheet, maybe a mistake?

Well, took your advice and used the other method, A(x) = pi x^(2) change in x for the disks.

(-x^(2)+1)^(2) == x^(4)-2x^(2)+1, after integrating I ended up with

pi * (1/5(x)^(5)-2/3x^(3) )+1x/ integrated from -1 to 1,

I ended up with (8/5)- (-8/5) *pi = 16pi/15 or 3.35.... correct

For the other method, I did
(x^(2)+1)^(2) - (2-0x)^(2) ... no
after multiplying and integrating this, i ended up with the equation,
1/5x^(5) - 2/3x^(3) -3x/ integrated from -1 to 1,

I ended up with -6.93*pi= -21.78. Not sure if this should automatically be an absolute value because its volume.

Thank you again, I'm just trying to understand and wrap my mind around this!
$\pi \int_{-1}^1 [(1-x^2) - (-1)]^2 - 1^2 \, dx$

$\pi \int_{-1}^1 (2-x^2)^2 - 1 \, dx$

$\pi \int_{-1}^1 3 - 4x^2 + x^4 \, dx$

.. now try it again.

also, understand that you should be making a graph sketch to determine which method is appropriate to determine the volume of revolution.

5. ## Re: I'm just incredibly frustrated right now with volumes of solid revolutions

Hey, thank you, I did try that second problem again and ended up with 56pi/15, hopefully this is correct. I am just having trouble visualizing these graphs in my mind, which my makes them hard to draw as well.

Thank you again.