solid, triangular base, square cross sections

[BobertGeraldo]

how would i set up the integral (to find volume) of a solid that had a base that was a right isoceles triangle. This triangle has sides of length L. And if you slice perpendicular to the base, but parallel to the hypotenuse, the cross section is a square. i think its like a pyramid.

i sorta drew a picture, with the y axis disectinf it down the middle, and i thought id integrate dy from 0 to L divided root 2. but im not sure what would be in the integral. Maybe (L divided by root 2)^2, since thats x and its a square. idk? any help would be greatly appreciated.

 

[skeeter]

let the base of the solid be bounded by the lines y = x, y = -x, and y = k, where \(\displaystyle k = \frac{L}{\sqrt{2}}\).

a slice perpendicular to the x-axis has side length s = (k - x) for x > 0.

so ... the volume of a representative cross-section is dV = (k - x)[supw9z3uhk]2[/supw9z3uhk] dx

using symmetry ... \(\displaystyle V = 2 \int_0^k (k - x)^2 dx\).

 

[soroban]

Hello, BobertGeraldo!

Set up the integral for the volume of a solid with a base that is a right isoceles triangle.
This triangle has sides of length L.
Cross-sections perpendicular to the base and parallel to the hypotenuse are squares.

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\(\displaystyle \text{The base is the isosceles right triangle bounded by: }\;y = x\text{ and }y = -x\)
. . \(\displaystyle \text{and the line: }\:x \:=\:\frac{L}{\sqrt{2}}\)

\(\displaystyle \text{The side of ths square is: }\:2y \:=\:2x\)
\(\displaystyle \text{The area of the square is: }\2x)^2 \:=\:4x^2\)

\(\displaystyle \displaystyle{\text{The integral is: }\;V \;=\;\int^{\frac{L}{\sqrt{2}}}_0 4x^2\,dx}\)