Question

calculus help

Answer 1

Write the integral in one variable to find the volume of the solid obtained by rotating the first quadrant region bounded by y=.5x^2 and y=x about the line x=3

Answer 2

If I help you on this one, I want to see you set up the problem.

Answer 3

ok first I have to find the radius. which would be (3-x), not sure about the intervals on this. and then I have to find the height. and the thickness

Answer 4

Oh wait do I need to use the disc method?

Answer 5

relate the radii of the solid to its height, dy, you're integral will look similar to

\[\int\limits_{0}^{h}f(y)dy\]

hint: h will be the intersection of
\[y=\frac{1}{2}x^2\] and \[y=x\]

Answer 6

Exactly. You are using the disk method for a rotation around a vertical line.

Answer 7

I set them equal to each other to find the intersection?

Answer 8

Or, solid of revolution I should say.

Answer 9

they intersect at (0,0) and (2,2)

Answer 10

Which one do you think is the height?

Answer 11

2?

Answer 12

Yep.

Answer 13

So you see how the solid is going to be hollow in the center?

Answer 14

ok i would use the equation: pi intergral a,b of [R(x))]^2dx

Answer 15

Close... your radius will need to be a function of y.... we are integrating the volume from y=0 to y=2

Answer 16

It would be very difficult to get a nice integral which relates the changing radius as we integrate in terms of x!

Answer 17

So replace the x with y and change dx to dy

Answer 18

so pi integral from 0 to 2 of [3(.5x^2)]^2 dy

Answer 19

you will have a inner radius which will swep around to create the empty space... and then an outer radius which is will sweep around as we rotate and create the surface. the total volume is the volume created by the outer surface minus the volume inside the inner surface

so our equation will have something to do with
\[R _{outer} (y) ^2 -R_{inner}(y)^2\]