Volume of solid bounded by surfaces

Question

Find the volume of the solid bounded by these two surfaces: $$ \text{surface a: }z=x^2+y^2\\ \text{surface b: }z=1-\sqrt{x^2+y^2} $$ Can someone please help me out? Should I use cylindrical or spherical? I've already tried both and can't find the correct answer! Thanks!

Answer 1

Cylindrical is the way to go (spherical will work, but when no spheres or expressions using $x^2+y^2+z^2$ appear, it's best to avoid them since they have messier formulae for things like $dV$ and $dS$). Here, we have no $x$ or $y$ separately, but $x^2+y^2$, which is a good hint to use cylindricals.

In cylindricals, the surfaces are $$ z = r^2, \qquad z=1-r. $$ First question: when do these intersect? When $r^2=z=1-r$, so $r^2+r-1=0$, or $r=\frac{-1 + \sqrt{1+4}}{2} = \frac{\sqrt{5}-1}{2}$ (the other root is negative, so meaningless since $r \geq 0$). Let's call this $a$ for short.

So the integral is over the region $r^2 < z < 1-r$, where $0<r<a$. $dV=r \, dr\, d\theta \, dz$ has no $z$-dependence, so we may as well do the $z$ integral first. Thus the volume is $$ V=\int_{r=0}^a \int_{z=r^2}^{1-r} \int_{\theta=0}^{2\pi} r \, d\theta \, dz \, dr. $$ This is simple enough: the inner integral gives $2\pi$. The $z$ integral gives $1-r-r^2$, so we have $$ V= 2\pi \int_{r=0}^a (1-r-r^2)r \, dr = 2\pi \left[ \frac{r^2}{2}-\frac{r^3}{3} - \frac{r^4}{4} \right]_0^a = \frac{a^2}{6}(6-4a - 3a^2 )\pi $$ Thankfully we know that $a^2=1-a$, so we can simplify this so we don't have to square $a$ directly: $$ \frac{a^2}{6}(6-4a - 3a^2 )\pi = \frac{1-a}{6}(6-4a - 3(1-a) )\pi \\ = \frac{(1-a)(3-a)}{6}\pi = \frac{3-4a+a^2}{6}\pi = \frac{4-5a}{6}\pi = \frac{13-5\sqrt{5}}{12}\pi $$

Answer 2

Using cylindrical polars $(r, \theta, z)$, we have the surfaces $S_a$ and $S_b$ given by $$S_a = \lbrace (r, \theta, z) : z = r^2 \rbrace$$ and $$S_b = \lbrace (r, \theta, z) : z = 1 - r \rbrace$$

Note in the region we want, $r^2 \leq 1-r$, so noting that $r^2 = 1-r \implies r = r_+ = \frac{\sqrt{5}-1}{2}$, we have $r \in [0, r_+]$.

Hence the volume is given by $$V = \int_0^{2 \pi} \int_0^{r_+} \int_{r^2}^{1-r} r \; \mathrm{d}z \, \mathrm{d}r \,\mathrm{d}\theta = 2 \pi \int_0^{r_+} (1-r)r-r^3 \mathrm{d}r = 2 \pi \left[ -\frac{r^4}{4} - \frac{r^3}{3} + \frac{r^2}{2} \right]_0^{r_+} = \frac{\pi}{6} r_+^2 \left( 6 - 4r_+ - 3r_+^2\right) = \frac{\pi}{12} \left(13 - 5\sqrt{5}\right) \approx 0.476$$