A hemispherical bowl of radius 5-cm is filled with water to within 3-cm of the top. Find
the volume of the water in the bowl?
$$\int_{0}^{2π}\int_{?}^{π}\int_{?}^{5}ρ^2\sinϕ\ dρdϕd\theta$$
What value should I put in the place of rho?
Find volume in spherical coordinates? $\iiint\rho^2\sinϕ\ dρdϕd\theta$
integrationmultivariable-calculusreal-analysisspherical coordinatesvolume
Related Solutions
Consider this figure which shows the two spheres and a plane at $z = 1$:
Find the volume of a "cap" of the yellow sphere (centered at the origin) above the blue plane. Multiply that volume by 2.0 due to symmetry, for the cap of the red sphere below the blue plane.
The differential volume of a thin slice disk of the yellow sphere is:
$\pi r^2$, where $r$ is measured perpendicular to the vertical ($z$) axis. Pythagoras tells us that $3^2 - z^2 = r^2$. (This incorporates your request to use polar coordinates.)
So your integral is
$\int\limits_{z=1}^3 \pi (3^2 - z^2) dz = {28 \pi \over 3}$.
But don't forget to multiply by 2.0 for the red cap beneath the blue plane:
$V = {56 \pi \over 3}$.
Your bowl can be modeled in 2-D by the equation $x^2+y^2=60^2$. We will make that in terms of $y$ to be $y=\sqrt{3600-x^2}$. We know the bounds are $50, 60$ because your shape is $10$ cm deep. In order to calculate the volume sweeped out by the curve between $50, 60$, we use the integral $$\pi\int_{50}^{60}\sqrt{3600-x^2}^2 dx$$ where the $\pi \sqrt{3600-x^2}^2 dx$ is the volume of a tiny cylindrical slice of your shape, and the integral is to add up those infinitelay many infinitely thin cylindrical slices into the shape the water forms in the bowl.
I think you can calculate the integral on your own.
Best Answer
In Cartesian coordinates, if you position the bowl such that it is given by equations $x^2+y^2+z^2 = 25$, $z\le 0$ (i.e. $z=0$ corresponds to the top of the bowl), the part of space taken by water can be described by conditions \begin{align} x^2+y^2+z^2 \le 25 & \quad (\text{the water is inside the bowl}) \\ z \le -3 & \quad(\text{up to within 3 cm from the top})\end{align} These inequalities written in spherical coordinates ($x=\rho\sin\phi\cos\theta$, $y=\rho\sin\phi\sin\theta$, $z=\rho\cos\phi$) take form \begin{align} 0 \le \rho \le 5 \\ \rho\cos\phi < -3 \end{align} From the latter you can conclude that $\cos\phi < 0$ , and you can rewrite the latter as $\rho > -\frac{3}{\cos\phi}$. So you have $$ -\frac{3}{\cos\phi} \le \rho \le 5$$ These conditions give you the limits of the integral over $\rho$. However, they also require that $ -\frac{3}{\cos\phi} \le 5$, or (remembering that $\cos\phi <0$) $$ \cos\phi \le -\frac35$$ $$ \phi \in [\arccos(-\frac35),\pi] $$ In conclusion the integral you need to calculate is $$ \int_0^{2\pi} \int_{\arccos(-3/5)}^\pi \int_{-3/\cos\phi}^5 \rho^2\sin\phi \,d\rho\,d\phi\,d\theta$$