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Calculus: Volume Problems

  • #1
1.Consider the given curves to do the following.
x=4+(y-3)[tex]^{2}[/tex], x=8

Use the method of cylindrical shells to find the volume V of the solid obtained by rotating the region bounded by the given curves about the x-axis. Sketch the region and a typical shell.

I'm lost on x being a function of y. How do you even enter these into a TI-83? Is there any way to make these easier?

Here's another one i've been working on that has caused me problems.
_________________________________________________________________
2.The region bounded by the given curves is rotated about the x-axis.
y=-x[tex]^{2}[/tex]+9x-18
y=0
Find the volume V of the resulting solid by any method.

First I graphed it:
htp://img377.imageshack.us/img377/3471/6338ol4.png (Just add another t in http)

cylindrical shell method
2 pi r h dr

I set up an integral. I get confused on determining the radius and height. If i'm rotating around the x-axis, i'm using y's. So, the radius should be y since it's centered around the y-axis. Then what is the shell height? The points at which the parabola crosses the curve are at x=3 and 6. So the shell height should be 6-y, but I think it should be where x = the equation.

But when I tried to single out x in the equation to get y as a function of x in y=-x[tex]^{2}[/tex]+9x-18, I couldn't calculate it.

Discs method
The area of one disc:
A(x)=[tex]\pi[/tex] * (-x[tex]^{2}[/tex]+9x-18)[tex]^{2}[/tex]

So the integral is
[tex]\pi[/tex] times the integral of (-x[tex]^{2}[/tex]+9x-18)[tex]^{2}[/tex] dx

Now the limits of integration should be from 3 to 6. I then integrated, plugged in the answer to my homework application which prompted me with a predictable "wrong" result. I've had no problems integrating, as i've completed 85% of my homework, but I have a hard time setting these problems up. Especially these odd-ball problems.

I'm sorry for my vague descriptions. It's hard to describe some of these things over the net. I appreciate any help. Thank you! :smile:
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Hi RedBarchetta! :smile:
I'm lost on x being a function of y. How do you even enter these into a TI-83? Is there any way to make these easier?
You can always change any variable letter into any other letter … provided, of course, you remember to change them back at the end!

Just interchange x and y, to give y = 4+(x-3)², y=8. :smile:
Then what is the shell height? The points at which the parabola crosses the curve are at x=3 and 6. So the shell height should be 6-y, but I think it should be where x = the equation.
Forget the word "height" … you need the area of each shell, and the thickness of each shell.

Then you integrate over the thickness … sometimes it's height, sometimes it's radius, sometimes … :confused:

In this case, the thickness is not x, but dx.

Just think of it as "thickness", and you won't be confused! :smile:
But when I tried to single out x in the equation to get y as a function of x in y=-x+9x-18, I couldn't calculate it.
Sorry … I don't understand this … y is a function of x. :confused:

Show us your working on the integral, and then we can see where the mistake is. :smile:
 
  • #3
Thanks for the help Tiny-Tim! I figured both of these out now. :approve:
 

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