Question

find the volume of the solid generated by revolving the region bounded by y=x^2 and y=sqrt(x) around the x-axis

Answer 1

find the volume created when you spin sqrt(x)

the subtract the volume you get when you spin x^2

the limits of integration being where the 2 curves meet

Answer 2

so far I have
\[x^2=\sqrt{x}\]
\[x^2-\sqrt{x}=0\]
\[x^2-x^{1/2}=0\]
\[2x-1/2x=0\]
\[x(2-1/2)=0\]
\[x=0, x=3/2\]
is this correct so far?

Answer 3

not quite;

if you can recall the shape of the graphs; these meet at only 2 points;

0^2 = sqrt(0)

1^2 = sqrt(1) are the only options available; so your math is errored someplace

Answer 4

is I wasnt supposed to integrate when I did

Answer 5

so x=0 and x=1 are the limits

Answer 6

yes, 0 to 1

recall the integrating a volume of rotation is adding up areas of circles:

pi f(x)^2

\[\int_{0}^{1}pi [x^{1/2}]^2dx - \int_{0}^{1}pi[x^2]^2dx\]

Answer 7

this cam be simplified; but its the same results

create a solid shape and dig out the part thats extra

Answer 8

The part I'm having trouble with is finding the limits

Answer 9

x^2 = sqrt(x) ; square the sides

x^4 = x

x^4 - x = 0

x(x^3 - 1) = 0

x = 0; x = 1

Answer 10

Ok now I think I understand

Answer 11

Thanks I am going to try and finish the problem myself.

Answer 12

good luck :)

Answer 13

thanks