Question

i have a problem.

Find the volume of the solid obtained by rotating the region bounded by the curves.
y=x^2 and y=1 about the line y=3

Answer 1

i used the washer method getting two radius (3-x^2) and (3-1)=2

\[\Pi \int\limits_{-1}^{1} (3-x ^{2})^{2}-(2)^{2}dx\]

i started of with this

Answer 2

y=x^2 and y=1 intersect at (0,0) and (1,1).
If the area is rotated about y=3, that would make many washers, therefore the washer method.

Take

Answer 3

Sorry, I though it was y=x, but it is y=1, so the limits are indeed from -1 to +1.

Answer 4

yea after i got the interval i factored out the two and got (x^4-6x^2+9) - (4) dx

Answer 5

then i took the integral of x^4-6x^2+5 and it came out to be x^5/5-6x^3/3+5x

Answer 6

after that i plugged in 1 and -1 and the answer i got was wrong

Answer 7

I don't get the same expression as you. It's (x^5-10x^3+25x)/10.
The numeric answer is 16/5.

Answer 8

multiplied by two pi to get 32pi/5.

Answer 9

i used washer method and got it wrong bt you used shell method...idk how to use shell for this one

Answer 10

I used the washer method also.
The integral is:
\[\int\limits_{-1}^{1}2\pi((1-x^2)(3-(1+x^2)/2)dx\]

Answer 11

whatttt?!?!! we never learned washer method with 2pi before

Answer 12

I suppose the 2pi has to come in whenever you rotate something, right?

Answer 13

yea whenever its been rotated bt we were given a formula like this
pi(outer radius)^2 - pi(inner radius)^2 dx

Answer 14

It actually comes to the same thing.
You would probably put:
\[\pi \int\limits_{-1}^{1}((3-x^2)^2-2^2)dx\]
and it gives the same answer of 32pi/5.

Answer 15

ohhhh ok thanks for the helpppp!!

Answer 16

You're welcome!