finding volume of a solid

[Guest]

I am not completely understanding the disk/washer method...

find the vol of the solid generated by revolving region bounded by the graphs of the equations about the given lines.
y=x^2 y=20x-x^2 a) x-axis b) the line y=102

for part a i am using the method, where R(x)=20x-x^2 and r(x)=x^2 and I get
volume=pi integral from 0 to 10 [(20x-x^2)^2-(x^2)^2]dx=104719.7551. is this right?

for part b i am also trying to use the washer method, where R(x)=102-x^2 and r(x)=102-(20x-x^2)^2.

vol=pi integral from 0 to 10 [(102-x^2)^2-(102-(20x-x^2)^2]dx but for some reason i am not getting what the answer is and I can't figure out why. I think it should be 108908.545324

can anybody help? Thanks

 

[galactus]

The first one is good. Fine work. Except write your answer as

\(\displaystyle \frac{100000{\pi}}{3}\) instead of the decimal form.

For about y=102. You appear to be correct. You're probably just making algebra errors.

\(\displaystyle \L\\{\pi}\int_{0}^{10}(102-x^{2})-(102-(20x-x^{2}))^{2}dx\)

\(\displaystyle \L\\{\pi}\int_{0}^{10}(40x^{3}-808x^{2}+4080x)dx=\frac{104000{\pi}}{3}\)

Leave your answer in pi form, not decimal form. For aesthetics if nothing else.

Shells will work, too, but it's trickier:

\(\displaystyle \L\\{2\pi}\int_{0}^{100}(y-102)(({10-\sqrt{100-y})-\sqrt{y})dy\)

You should get the same as with the washer method.

 

[Guest]

thanks for help. good to know that i am on the right page, excluding a few calculating errors. my professor didn't explain these methods very well so i am finding it a little difficult to fully understanding.

thanks again!