volumes of revolution / centroids

johnboy

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Jul 20, 2006
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Hi guys! I just took this quiz and I was wondering if you could tell me if my answers are right or not. (Please, God: let them be!)

Consider the region R, which is bounded by the x-axis, y = x, and x + y = 4.

A) What is the volume when rotating R about the x-axis?

I did the formula (pi) integral R^2 - r^2 dx and got 16 (pi).

B) What is the volume when rotated about x = -2?

This time, I found the equations with respect to y, and got 32 (pi).

C) Find the centroid of R.

I did "1/A integral xf(x)dx" to get the X Moment, for which I got 2/3. Then I did "1/2X integral f(x)^2 dx" and got the Y Moment of 8/3.

Thanks so much guys!!
 
Did you graph these lines?
It seems to me as if you have the wrong region R.
 
ah, i really hope i did it right... :-/

When i graped it, i got a triangle that had its heighest point at (0,4) and the vertex at (2,2)... ??
 
But that triangle is bounded by the y-axis not the x-axis.
 
Hello Johnboy:

Here's the region you're revolving. Try taking it in 2 subregions.

\(\displaystyle y=4-x\;\ y=x\;\ \text{about x-axis}\)

Try it with disks, then shells. See if both methods give the same result.

vorpe9.jpg


disks:

\(\displaystyle \L\\{\pi}\int_{0}^{2}{x^{2}}dx+{\pi}\int_{2}^{4}(4-x)^{2}dx\)

I believe you tried \(\displaystyle \L\\{\pi}\int_{0}^{2}((4-x)^{2}-x^{2})dx\)

You end up with \(\displaystyle \L\\{\pi}\int_{0}^{2}(8x-16)dx\)

Not the integral you need.
 
johnboy said:
ah! will that screw up all my answers..?
Well that depends on what the question said.
Did it say 'bounded by the x-axis'?
Oe did it say 'bounded by the y-axis'?
 
it said bounded by x... i was just rushing and not thinking :-( Wow, im a freaking idiot. This is great.
 
johnboy said:
it said bounded by x... i was just rushing and not thinking :-( Wow, im a freaking idiot. This is great.

No you are not. Just rushing is all which we all do it one time or another. As long as you learn from it is ok. Take it from a guy named Jonboy. ;)
 
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