Question

Volumes of rotation (I think thats what this is called?)

Answer 1

\[\pi \int\limits_{0}^{1}(y^3)^2dy=\frac{ \pi }{ 7 }\]

But thats not an answer choice.
Is it - pi/7 and if so why?

Answer 2

Hey

Answer 3

hello

Answer 4

yes

Answer 5

When we revolve a sharp around x axis we use the formula
Volume = π∫y²
Agreed?

Answer 6

right

Answer 7

So when we are going to revolve a shape around y axis the formula will change to
Volume = π∫x²
Agreed?

Answer 8

oh. lol I see where youre going with this

Answer 9

I need to solve for x

Answer 10

wait

Answer 11

Oh no no, fortunately they have give you the equation of the curve in the form of x

Answer 12

yea. you're right

Answer 13

The equation of the curve is x = y³
When rotated around the y axis, the volume will equal to
Volume = π∫x²
Volume = π∫(y³)²
Can you work it out?

Answer 14

Did you sketch it? That seems to be your problem. You got the wrong region.

Answer 15

The method I'm seeing is to find the area between two "curves" which is the y = (0,1) area of x= 1 and x=y^3

Answer 16

@Lord_Box We just will take y =1 and y=0 as limits in this integral. That's same as to what you are implying

Answer 17

http://www.wolframalpha.com/input/?i=region+bounded+by+x%3Dy%5E3,+x%3D1,+y%3D0

Outer radius is x=1, inner radius is x=y^3, thickness is dy. Use outer minus inner
\[\large \pi \int\limits_{0}^{1}(1 - y^3)dy\]

Answer 18

\[\pi \int\limits_{0}^{1}(y^3)^2=\pi \int\limits_{0}^{1}y^6=\pi[\frac{ y^7 }{ 7 }]_{0}^{1}=\pi[\frac{ 1 }{ 7 }-0]=\frac{ \pi }{ 7 }\]

Answer 19

@agent0smith The method you gave will give anwer 3pi/4

Answer 20

Not drawing a sketch to make sense of things is near criminal in a problem like that (i mentally sketched it in my head)