# Thread: Volume of solid generated by curves. ( I tried my best!)

1. ## Volume of solid generated by curves. ( I tried my best!)

Q. The region enclosed by the given curves is rotated about the specified line. Find the volume of the resulting solid.

y=x^3 and y=sqrt.of(x); about x = 1.

What I did: Rotating about a vertical line along the y-axis.
x^3=sqrt.of(x); x= 0 and 1
x=y^(1/3) and x=y^2.
=pi(1+y^(1/3))^2-pi(1+y^2)^2
V= pi*Integral from 0 to 1 of [(1+(y^1/3))^2-(1+y^2)^2]dy
V= pi*Integral from 0 to 1 of [(1+y^(2/3)+2y^(1/3)-(1+y^4+2y^2)]dy
V= pi[(3y^(2/3))/2 + (6y^(4/3))/4 - (y^4)/4 - (2y^3)/3] from 0 to 1

But after plugging in for 0 and 1, I got 25pi/12. But the ans. is 17pi/30...8(

2. but that's not the answer though. Plus are you able to do it my way?

3. I wonder why you put 1+y^{1/3}... But since I never worked with solid or rotation other than about the main axes, I would have gone with this:

From what I see, you need to move both curves to the left by one unit.

Hence, the equation of the new cubic becomes: $y = (x+1)^3$

And that of the square root becomes: $y = \sqrt{x+1}$

The volume then becomes:

$\displaystyle \pi \int^1_0 (1-y^2)^2 - (1-y^{\frac13})^2\ dy$

Which gives:

$\displaystyle \pi \int^1_0 y^4-2y^2 - y^{\frac23}+2y^{\frac13}\ dy$

$\displaystyle \pi \left[\dfrac15y^5 - \dfrac23y^3 - \dfrac35y^{\frac53}+\dfrac32y^{\frac43}\right]^1_0$

$\displaystyle \pi \left[\left(\dfrac15 - \dfrac23 - \dfrac35+\dfrac32 \right) -(0) \right]^1_0$

Are you sure the answer is 17pi/30? I'm getting 13pi/30...

Or I missed something...

4. method of cylindrical shells w/r to x ...

$\displaystyle V = 2\pi \int_0^1 (1-x)(\sqrt{x} - x^3) \, dx = \frac{13\pi}{30}$

method of washers w/r to y ...

$\displaystyle V = \pi \int_0^1 (1 - y^2)^2 - (1 - \sqrt[3]{y})^2 = \frac{13\pi}{30}$

5. you guys are right. It's 13pi/30. Thanks a lot. Some dumb mistakes huh?