# Math Help - Rotate About x-axis

Find volume of solid generated by revolving the region bounded by the graphs of the equations about the indicated lines.

y = x^2, y = 4x-2x^2

I set x^2 = 4x-2x^2 to find my limits of integration.
I found my limits to be [0,2].

2. ## Re: Rotate About x-axis

Originally Posted by nycmath
Find volume of solid generated by revolving the region bounded by the graphs of the equations about the indicated lines.

y = x^2, y = 4x-2x^2

I set x^2 = 4x-2x^2 to find my limits of integration.
I found my limits to be [0,2].
$x^2=4x - 2x^2$

$3x^2 -4x =0$

$3x\left(x-\frac 4 3 \right)=0$

$x=0 \vee x=\frac 4 3$

3. ## Re: Rotate About x-axis

I made a mistake. The correct functions are
y = x^2, y = 4x-x^2.
Now, we can say that the limits of integration are [0,2].

x^2 = 4x-x^2

2x^2-4x = 0

2x(x-2) = 0

Solving for x, I get x = 0 and x = 2.
How can the volume be 32pi/3?

4. ## Re: Rotate About x-axis

Originally Posted by nycmath
I made a mistake. The correct functions are
y = x^2, y = 4x-x^2.
Now, we can say that the limits of integration are [0,2].

x^2 = 4x-x^2

2x^2-4x = 0

2x(x-2) = 0

Solving for x, I get x = 0 and x = 2.
How can the volume be 32pi/3?
That's what the answer is. Just find

$\pi \displaystyle{\int_0^2}\left(4x^2-x^2\right)^2-\left(x^2\right)^2~dx$

do you understand this?

5. ## Re: Rotate About x-axis

Yes, it is the top function squared minus the bottom function squared.

I can take it from here.

Tell me, what if we are given y = sqrt(x),
y = 0, x = 4 and told to rotate about the y-axis? What is the volume in this case?

6. ## Re: Rotate About x-axis

Originally Posted by nycmath
Yes, it is the top function squared minus the bottom function squared.

I can take it from here.

Tell me, what if we are given y = sqrt(x),
y = 0, x = 4 and told to rotate about the y-axis? What is the volume in this case?
$2\pi \displaystyle{\int_0^4}x \sqrt{x}~dx$

$\sqrt{x}\geq 0~~\forall x$ so I can omit the absolute value sign.

7. ## Re: Rotate About x-axis

Where did 2pi come from?