Question

I kinda know how to do this...

Answer 1

you will get a cone of base radius = 3 and height = 1
volume = 1/3*pi*r^2*h

Answer 2

how'd you do that?

Answer 3

Oh boy, I remember doing this lol, calculus?

Answer 4

spin that R1 region around x axis
what region does it cover in 3D ?

Answer 5

Yes I dislike this very much

Answer 6

mmm ok I can see that,but it is asking about the specific line I didn't think the x-axis I thought it was the line labeled on the graph.

Answer 7

calculus is like icecreams and cheesecakes and i see you're getting into the exciting part of it, you will like it sooner or later im sure

Answer 8

OA is on x axis

Answer 9

wait i see it now i didn't read it right

Answer 10

ok so what about the R2 region about the AB... I can see the cone with out drawing it but this one i can't form a picture.

Answer 11

as a start, draw a rough sketch, label whatever you think is important and draw a "volume element"

Answer 12

\[r = x~~~ \int\limits_{a}^{b} \pi r^2 dx = \pi \int\limits_{0}^{1} (x)^2 dx = \pi (1/3x^3) ~~ from ~~0~~\to~~1 = \pi/3\] using the disk/ washer method someone can check this if it's correct, been some time since I've done this :P, remember we're integrating along the axis parallel to the axis of rotation.

Answer 13

Because we're rotating R1 about OA

Answer 14

for R1 around OA, we should get 3pi right ?

Answer 15

still doing R1 around OA :

volume element = pi*y^2*dx = pi*(3x)^2*dx

Answer 16

Was doing \[y=\sqrt[4]{x}\]

yeah I think you're right, this damn post button isn't working...

Answer 17

yeah refresh the page when it hides under the chat buttons