Let $z = 1/r^n$, where $r = \sqrt{x^2+y^2}$ and $0 < n < 2.$ Note that this function has a discontinuity at the origin. Find the volume, $V(a)$, under this surface (and above the xy-plane) over the annulus: $a ≤ r ≤ 1$, where $0 < a < 1.$
This question has completely thrown me for a loop. I understand that this SHOULD be a triple integral, however, I can't figure out what to do with the n. It seems to me that this should also be integrated, however, this would imply a quadrupled integral, which we have not used in class and seems very unusual to introduce. Where am I going wrong here?
Since $r \ge 0$, $z = r^{-n} \ge 0$. Thus the volume is $$V(a) = \iint_A (\sqrt{x^2+y^2})^{-n}$$ Where $A$ is the annulus $a\le r\le 1$. Applying polar coordinate transforms to this gives $$V(a) = \int_0^{2\pi} \int_a^1 r \cdot r^{-n}\ \mathrm dr\ \mathrm d\theta = \frac{2\pi}{2-n} [r^{2-n}]_a^1 = \frac{2\pi}{2-n} (1-a^{2-n})$$ Assuming $n\ne 2$. If $n=2$ we get $$V(a) = 2\pi \int_a^1 \frac1r \ \mathrm dr = -2\pi \ln a$$