mx2 volume question (1 Viewer)

[CcYann]

the base of a particular solid is x^2+y^2=4. find the volume of the solid if every cross-section perpendicular to the x-axis is a parabolic segment with axis of symmetry passing through the x-axis and height the length of its base.

I have drawn the diagram for it, but I don't know how can I calculate the area/volume of the parabolic segment.
any ideas for it?

thanks in advance

 

[bleakarcher]

Let the segment taken at some arbitrary distance x from the origin (where x lies in the interval [-2,2]) have volume δV and thickness δx.
By Simpson's law:
δV=(8/3)y^2*δx (note it is only approximately equal to this expression)
V=sum from x=-2 to x=2 of [(8/3)y^2 * δx]
However, this is only an approximation for the true volume of the solid. As δx->0, the slices become smaller and smaller and therefore we get closer and closer to the true volume.
so
the true volume V=lim[δ->0] sum from x=-2 to x=2 of [(8/3)y^2 * δx] and you can now find the volume using the definition of the integral.

Hope this helped

 

[CcYann]

Let the segment taken at some arbitrary distance x from the origin (where x lies in the interval [-2,2]) have volume δV and thickness δx.
By Simpson's law:
δV=(8/3)y^2*δx (note it is only approximately equal to this expression)
V=sum from x=-2 to x=2 of [(8/3)y^2 * δx]
However, this is only an approximation for the true volume of the solid. As δx->0, the slices become smaller and smaller and therefore we get closer and closer to the true volume.
so
the true volume V=lim[δ->0] sum from x=-2 to x=2 of [(8/3)y^2 * δx] and you can now find the volume using the definition of the integral.

Hope this helped

the final answer is correct, but how can u get (8/3)y^2 as the segment's area?

 

[bleakarcher]

sorry, I thought you knew how to. You use Simpsons rule.

 

[CcYann]

sorry, I thought you knew how to. You use Simpsons rule.

oh, dumb me. thks a lot.