Disks or shells? x^2 = 8y, y = 4, and plane sections....

[tyronne]

Find the volume of the solid whose base is the region bounded by the parabola x^2=8y and the line y=4, and each plane section perpendicular to the y-axis is an equilateral triangle. (which has an area of sqrt(3)/4 * s^2 i believe)

I've gone through this many times now and I can't seem to set the integral up properly, using both shells and disks...
Please help me

 

[Deleted member 4993]

Re: Disks or shells?

Find the volume of the solid whose base is the region bounded by the parabola x^2=8y and the line y=4, and each plane section perpendicular to the y-axis is an equilateral triangle. (which has an area of sqrt(3)/4 * s^2 i believe)
I've gone through this many times now and I can't seem to set the integral up properly, using both shells and disks...
Please help me

Please show your work - even if you know it is wrong - so that we know where to begin to help you.

 

[soroban]

Hello, tyronne!

Find the volume of the solid whose base is the region bounded by the parabola \(\displaystyle x^2=8y\)
and the line \(\displaystyle y=4\), Each plane section perpendicular to the y-axis is an equilateral triangle.
(which has an area of sqrt(3)/4 * s^2 i believe)

I've gone through this many times now and I can't seem to set the integral up properly,
using both shells and disks. . . . . There is your problem!

"Shells and disks" are used when a region revolved about the line.
. . This is not a solid of revolution.

This is a "Cross-section" problem.

Find the area of the cross-section, multiply by the "thickness".
Then integrate and evaluate the resulting expression.

 

[galactus]

If you let A(y) equal the area of the equilateral triangle whose sides are length 2x, then

\(\displaystyle A(y)=\frac{\sqrt{3}}{4}(2x)^{2}\)

But \(\displaystyle x=\sqrt{8y}\)

\(\displaystyle A(y)=\frac{\sqrt{3}}{4}(2\sqrt{8y})^{2}\)

Now, finish?.