A region R, in the first quadrant only, is enclosed by y=x^3 and y=cubed root of x. find the volume of the solid obtained by revolving the region R about x=-2.

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- anonymous

- katieb

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- anonymous

First you have to imagine and draw a rough of what these would look like. You would find examples of these functions in the first chap of a calculus book or put them in a calculator

- anonymous

okay i imagined it and drew a picture. what do i do next?

- anonymous

Then you find the boundaries of integration (it might be obvious, it starts at 0) but set x^3=cube root of x; solve for x. The points tell you where y=x^3 and y=cube root x intersect (boundaries of integration)

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## More answers

- anonymous

ehhh i tried but i dont know how to find the boundaries. i got x^3-sq root of x=0 but i dont know what that would make x equal?

- anonymous

x^3=cube root of x
The problem is the cube root sign. Just cube each side of the eq
(x^3)^3=(cube root of x)^3

- anonymous

okay i got x^9-x^2

- anonymous

how do i factor that?

- anonymous

Good. but you are off a little. It should be
x^9-x=0
Both x^9 and x have a common factor of x, so you have to pull it out

- anonymous

but if i square the squareroot of x, isnt is x^2?

- anonymous

Actually if you square the square root of x, you get x; likewise cube the cube root you get x

- anonymous

ohhhh okay sorry i got it!

- anonymous

So, you have a lot of catching up, I'm going to go fast because I have to go. So the boundaries are 0 to 1

- anonymous

okay next?

- anonymous

You are going to be using washer method

- anonymous

Radius of big washer is x^3 +2
radius of small washer is cube root of x +2

- anonymous

okay ill try that!

- anonymous

Integrate from 0 to 1 [{(x^3+2)^2}pi - {(cube rooot x +2)^2}pi]

- anonymous

okay i will do that now. whats the final answer?

- anonymous

Haven't work it out. Setting it up is the fun part. Integrating is the dirty work. I left the dirty work for you.

- anonymous

okay thanksss for all the help!

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