I have this seemingly simple volume of a solid of revolution, but the limits and function are unknown.

Question

How can I possibly find the numerical area of the region without knowing the function itself or the limits?

Answer 1

Using cylindrical coordinate system $$V=\int_{0}^{2\pi}\int_a^b\int_{0}^{f(r)}rdzdrd \theta=2\pi \int_{a}^{b}f(r)rdr$$

We know the result is $2\pi$, therefore $$\int_{a}^{b}f(r)rdr=1$$

If the volume is made by revolving about $x=-1$, you can do similar thing but just shift the coordinate by -1. $$\int_{a+1}^{b+1}f(r+1)rdr=4$$

By substituting, you get $$\int_{a}^{b}f(r)rdr-\int_{a}^{b}f(r)dr=4$$

Thus we get the area $$\int_{a}^{b}f(r)dr=-3$$ Interesting enough, it seems that the graph in the book tries to make you think the area is positive.

Answer 2

Pappus' (second) centroid theorem is of help here. (I don't recognize the book you are taking this problem from, so I don't know if you've covered this topic.) The volume of a solid of revolution is given by $ V \ = \ 2 \pi \ r_c \ A \ $ , where $ \ r_c \ $ is the perpendicular distance of the centroid of the region being revolved from the axis of revolution.

Let the coordinates of the centroid be $ \ (\overline{x} \ , \ \overline{y} ) \ $ . The volume of the solid produced by revolution about the $ \ y-$ axis is then

$$ V_y \ = \ 2 \pi \ \overline{x} \ A \ \ = \ \ 4 \pi \ \ , $$

while the volume of the solid produced by revolution about the vertical axis $ \ x \ = \ -1 \ $ , which is one unit further to the left, is

$$ V_{vert} \ = \ 2 \pi \ ( \overline{x} \ + \ 1) \ A \ \ = \ \ 8 \pi \ \ . $$

We can thus locate the $ \ x-$ coordinate of the centroid of the region $ \ S \ $ as $ \ \overline{x} \ = \ 1 \ $ , which thereby tells us that the area of said region is $ \ A \ = \ 2 \ $ .