Would anyone please help me solve and explain this question. This question was one of the questions on my calc book that didn't have an answer nor a solution for. I want to know how to do it, but the book is not offering that (unless I buy the solution key, which I can't afford). I hope you guys can help; here is the question
Find the volume of the wedge by integrating the area of vertical cross sections. Note that vertical cross sections are right triangles the height and base of which depend on how far the cross section is from the origin.
Please forgive me if I missed something out, I am new to this website and I just wish someone could help with this problem.
Update: Here is what I did:
The line from 4 to 8 is given by the equation (z/4)+(x/8) = 1 and the line
from 8 to 6 is given by (y/6)+(x/8) = 1. The cross sections perpendicular
to the x-axis are right triangles with height 4(1 - x/8) and base b(1 - x/8).
Thus, we have:
Integral (1/2)4*6(1 - x/a)^2 dx (with limit = 0 to limit = 8) = 1/6 * (6*4*8) = 32
What I am not very sure about is the first steps before taking the integral! Would any one explain it further in details, and relate it to how we get the answer? I hope that adds to the question.
$$
S : \frac{x^2}{4} + \frac{y^2}{9} = 1
$$
Because the cross sections are isosceles right triangles with hypotenuse in the base, the length of the hypotenuse is $ 6\sqrt{1-\frac{x^2}{4}}$, which means that the area of the cross section is $9(1-\frac{x^2}{4})$
Best Answer
Project the moving triangle on the $yz$ plane. This projection preserves areas.
We are now faced with a 2D issue with a moving line. The base triangle has area 12.
The ratio of the area of the moving triangle and the base triangle is equal to the "shrinking" factor
$$f(x)=\left(\dfrac{8-x}{8}\right)^2$$
(for example, if $x=4$, the surface is shortened by a factor $f(4)=1/4$).
It remains to compute the volume of the wedge (or tetrahedron) as the integral of the area of its sections:
$$\int_0^8 12 f(x) dx = 32.$$