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anonymous
 5 years ago
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.
anonymous
 5 years ago
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
 5 years ago
Best ResponseYou've already chosen the best response.0First 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
 5 years ago
Best ResponseYou've already chosen the best response.0okay i imagined it and drew a picture. what do i do next?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then 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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ehhh i tried but i dont know how to find the boundaries. i got x^3sq root of x=0 but i dont know what that would make x equal?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^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
 5 years ago
Best ResponseYou've already chosen the best response.0how do i factor that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good. but you are off a little. It should be x^9x=0 Both x^9 and x have a common factor of x, so you have to pull it out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but if i square the squareroot of x, isnt is x^2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually if you square the square root of x, you get x; likewise cube the cube root you get x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohhhh okay sorry i got it!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, 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
 5 years ago
Best ResponseYou've already chosen the best response.0You are going to be using washer method

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Radius of big washer is x^3 +2 radius of small washer is cube root of x +2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Integrate from 0 to 1 [{(x^3+2)^2}pi  {(cube rooot x +2)^2}pi]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay i will do that now. whats the final answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Haven't work it out. Setting it up is the fun part. Integrating is the dirty work. I left the dirty work for you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay thanksss for all the help!
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