The problem involving cross sections
I am so confused on how to find volume using known cross sections. I've never understood it. This problem that I've encountered is very difficult, and I tried using the formula of the area of an isosceles triangle, but I don't know where to go from there. I would be so grateful if someone could help.
Let $f(x) = 4\cos \left( \frac{\pi x}{4} \right)$ and $g(x) = (x-2)^2$
Our cross sections are isosceles right triangles, so our legs are the same length, and it is given that a leg lies in $R$ and is perpendicular to the $x$ axis.
The area of a isosceles right triangle is $\frac12 \ell^2$ where $\ell$ is the length of the leg.
Therefore the area of the isosceles right triangle is $ \frac12 (f(x) - g(x) )^2 = \frac12 \left((x-2)^2 -4\cos \left( \frac{\pi x}{4} \right) \right)^2$
Now we integrate from $x=0$ to $x=2$ as shown:
$$\int_0^2 \frac12 \left((x-2)^2 -4\cos \left( \frac{\pi x}{4} \right) \right)^2 dx$$
Evaluating the integral by hand isn't pretty, but certainly doable, and through WolframAlpha we see that this integral simplifies down to about $1.775$