Question

Find the volume of the solid obtained by rotating the region bounded by
y = 2x^2 and y = 3 and x = 0 about the y-axis.

Answer 1

i think this is just
\[\pi\int_0^3\frac{y}{2}dy\] malevolence?

Answer 2

Your region is only the left side.
http://www.wolframalpha.com/input/?i=plot+y%3D2x^2%2Cy%3D3
Set 2x^2=3
x=+/-sqrt(3/2)
Use 0 to the positive one.
Using shell method:
The height is 3-2x^2; the radius is x
You have:
\[2 \pi \int\limits_0^{\frac{\sqrt3}{2}} x(3-2x^2)dx\]
Which gives \[\frac{9 \pi}{4}\]

Answer 3

Should read "region is on right side"

Answer 4

sole for x and get
\[x=\sqrt{\frac{y}{2}}\] then use
\[\int \pi r^2\] yes?

Answer 5

hold the phone. you are rotating around y - axis yes?

Answer 6

Yes, and you are given y=2x^2 y=3 so you should use shell method so you can slice parallel to the axis of rotation.

Answer 7

satellite, we get the same answer, you just used a different method :P

Answer 8

yes or you can use
\[\pi r^2\] get the same thing. i think integrating wrt y is easiest here maybe

Answer 9

you are right. get the same thing

Answer 10

http://www.wolframalpha.com/input/?i=integral+2pi*x%283-2x^2%29%2Cx%2C0%2Csqrt%283%2F2%29
http://www.wolframalpha.com/input/?i=integral+pi*y%2F2%2Cy%2C0%2C3

Answer 11

You're probably right, I just saw a x^2 and was like "solving for x gives a pm" so I went with shell

Answer 12

got it. here is a picture of the region to be rotated. that is why i used that method
http://www.wolframalpha.com/input/?i=y+%3D+2x^2+%2C+y+%3D+3+%2C+