Question about using Washer method

Question

I want to use the Washer method of integration to find the volume of material in a paraboloid of thickness 0.5mm. For the Washer equation $$V= π∫b[f(x)^2−g(x)^2]dx$$ $f(x)$ is a parabola, and I know that $g(x)$ is a similar parabola that is 0.5mm 'in' from $f(x)$, if that makes sense. The parabola is cut off with a plane at $x=b$. I suppose I could find the volume of $f(x)$ and $g(x)$ with disc method and subtract $g(x)$ from $f(x)$, or do Washer method. Would it be a valid method to use the Washer equation with $g(x)$ simply being whatever $f(x)$ is translated right by 0.05 cm? If not, how would you do it?

Answer 1

The integral method is fine as long as you have the correct functions $f(x)$ and $g(x)$.

However, if the paraboloid maintains the same thickness everywhere, you can not simply shift $f(x)$ right by d and then use it for $g(x)$. Such shift only guarantees the thickness d at the vertex, while wrong elsewhere. In fact, the wall would become thinner as $x$ moves to the right.

Furthermore, $g(x)$ is not be a parabola. For small thickness, though, you may derive a good approximation for $g(x)$.

You may proceed with the derivation by finding the corresponding $(x’,y’)$ on $g(x')$ given $(x,y)$ on $f(x)$. For the fixed thickness d, they are related by, assuming $f(x)=\sqrt{x}$,

$$x’=x+\frac{d}{\sqrt{1+4x}}\\ y’=y-\frac{2d\sqrt{x}}{\sqrt{1+4x}}$$ Next, replacing $y$ with $x^2$ and then eliminating $x$, you find the equation relating $y’$ and $x’$ and, in turn, $g(x’) = y’(x’)$. As can be seen, $g(x)$ is rather complex; obviously, it is not a parabola.