The base of a solid in the first quadrant of the xy plane is a right triangle bounded by the coordinate axes and the line x + y

Question

4 years ago

15

The base of a solid in the first quadrant of the xy plane is a right triangle bounded by the coordinate axes and the line x + y

= 2. cross sections of the solid perpendicular to the base are squares. what is the volume, in cubic units, of the solid?

Mathematics

1 answer:

Ivahew [28] · Answer 1 · 2021-12-31T12:30:26+03:00

The area of an equilateral triangle of side "s" is s^2*sqrt(3)/4. So the volume of the slices in your problem is

(x - x^2)^2 * sqrt(3)/4.

Integrating from x = 0 to x = 1, we have

[(1/3)x^3 - (1/2)x^4 + (1/5)x^5]*sqrt(3)/4

= (1/30)*sqrt(3)/4 = sqrt(3)/120 = about 0.0144.

Since this seems quite small, it makes sense to ask what the base area might be...integral from 0 to 1 of (x - x^2) dx = (1/2) - (1/3) = 1/6. Yes, OK, the max height of the triangles occurs where x - x^2 = 1/4, and most of the triangles are quite a bit shorter...