Volume of Solid of Revolution with Disk Method: y=5x^2, x=1, y=0 about x-axis

[Bricenob]

I was wondering if anyone could help me with this. I'm stuck and not sure where to start/how to go about it and finding the integral as well...

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=5x^2,x=1,y=0, about the x-axis

 

[HallsofIvy]

Imagine a line drawn perpendicular to the x-axis up to the give parabola. Rotated around the x-axis, it will sweep out a disk with area $\pi y^2$ so, imagining a thickness of "dx", volume $\pi y^2 dx$. Integrate that from 0 to 1.

 

[Bricenob]

Imagine a line drawn perpendicular to the x-axis up to the give parabola. Rotated around the x-axis, it will sweep out a disk with area $\pi y^2$ so, imagining a thickness of "dx", volume $\pi y^2 dx$. Integrate that from 0 to 1.

so when I do that.. I come up with (pi)(int 0-->1)(5*((x^3)/3)= (pi) (int. 0-->1) 5(1^3)/3= 5pi/3. But it says this answer is incorrect. Where am I going wrong?

 

[HallsofIvy]

If $y= 5x^2$ then $y^2$ is NOT equal to "$5x^3/3$"!

Or did you mean that $5x^3/3$ is the result of integrating $x^2 dx$? (If so do NOT continue to write "int. 0-->1"!)

That would be the integral of $\pi x^2dx$ but that is NOT what you should be integrating! You should be integrating $\pi y^2 dx$. And, of course, $y= 5x^2$.

 

[phion]

I think the disk method in terms of x should work.