Volumes of Solids Shell, Disks and Washers

[shiela.bxoxo]

[h=1]Volume of Solids Shell, Disk, and Washer method.[/h] I'm stuck on these problems. If you can explain the steps, please and thank you!


1.)Using any method
The region bounded above by the line y = 16, below by the curve y = 16-x^2, and on the right by the line x = 4 about the line y = 16

2.) Using Shell
Volume of the region left of x = y+6 and below y^2 and above the x-axis. About y=3.

Any method
3.) volume of the region below y= -3x+6 and enclosed by the y-axis from 0 to 2 about the x-axis.

4.) the volume of the region left of y = sqrt(2x) and below y = 2 about the y-axis.

5.) A solid lies between a plane perpendicular to the x-axis at x = -6 and x = 6. The cross sections perpendicular to the x-axis between these planes are squares whose diagonals are from the semicircles y = -sqrt(36-x^2) to the semicircle y = sqrt(36-x^2)

 

[tkhunny]

These are fine problems. What have you tried? What were the results? Did you try both methods?

I like to ponder this, based on partial derivatives:


\(\displaystyle 2\cdot\pi\int r\cdot h\;dr\)


\(\displaystyle \pi\int r^{2}\;dh\)

This helps me visualize what's what.

 

[shiela.bxoxo]

These are fine problems. What have you tried? What were the results? Did you try both methods?

I like to ponder this, based on partial derivatives:


\(\displaystyle 2\cdot\pi\int r\cdot h\;dr\)


\(\displaystyle \pi\int r^{2}\;dh\)

This helps me visualize what's what.

Yes I'vve tried the washer method, I'm just confused on how to assemble it in the beginning.

 

[tkhunny]

Well, what did you get with your washer method. That second one is the one you need. There's no hole in the middle, so "washer" is a bit of a misnomer. Maybe a disk?

\(\displaystyle \pi\cdot\int r^{2}\;dh\)

What's the radius of each disk? You should get 16-y. Stare at your graph until you see it. You did draw it, right?

What is the height of each disk? This is where the calculus invites you in. dx

So far, we have this:

\(\displaystyle \pi\cdot\int (16-y)^{2}\;dx\)

Now what?

 

[shiela.bxoxo]

Well, what did you get with your washer method. That second one is the one you need. There's no hole in the middle, so "washer" is a bit of a misnomer. Maybe a disk?

\(\displaystyle \pi\cdot\int r^{2}\;dh\)

What's the radius of each disk? You should get 16-y. Stare at your graph until you see it. You did draw it, right?

What is the height of each disk? This is where the calculus invites you in. dx

So far, we have this:

\(\displaystyle \pi\cdot\int (16-y)^{2}\;dx\)

Now what?

Yes, I used the disk method. They're about the same. Second radius being zero
Anyways, so I integrated it and got the answer, but it is answer choice right? and the answer I got werent in the answer choices, I was thinking maybe I did it wrong. Maybe subtract the radius by 6 or add?

I get 128pi which is not in the answer choices.

 

[tkhunny]

You're just not going to show any work, are you?

128pi is actually not too bad. It is a reasonable estimate for checking magnitude. But, no, that's not good. It's quite a bit bigger than that.

Having said that, answer my question. What's next? We have "dx" and the argument is in terms of 'y'. That's no good. We need a litle algebra.