Find volume in spherical coordinates? $\iiint\rho^2\sinϕ\ dρdϕd\theta$

Question

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?

Answer 1

If you have to use spherical coordinates to find this volume (so I am intentionally not going to discuss other approaches), then your limits of integration are definitely incorrect. One important observation is that the bounds for $\rho$ can't both be constants, because the shape is NOT a spherical shell, i.e. it's NOT confined between two spheres $\rho=\text{constant1}$ and $\rho=\text{constant2}$.

How exactly you set up such a triple integral depends on how your position the bowl in the space. I think it would be easiest to place the center of the bowl at the origin with the bowl being below the $xy$-plane. See the picture below for a sideways view — the bowl is outlined in red, and the green line is the water level.

First, let's determine the bounds for $\varphi$. Note that with the bowl positioned like this, $\varphi$ does not start at $\varphi=0$; instead, it starts at the angle shown by the upper dashed radius and continues down to $\varphi=\pi$. What is that initial angle? Note that it's determined by the line that goes to the point "$(4,-3)$" (in this projection on the picture, not in the actual 3D space), so this angle can be described as $\varphi=\pi-\arctan(4/3)$.

Now, the bounds for $\rho$. Look at the other generic dashed radius. It shows that $\rho$ goes from the green line on the picture, which in this sideways view represents the horizontal plane $z=-3$, to $\rho=5$ when it hits the sphere. Converting the equation of the plane $z=-3$ to spherical coordinates, we get $\rho\cos\varphi=-3$ or $\rho=-3\sec\varphi$.

I hope you can set up the triple integral now.

Answer 2

The water is 2cm deep, so forms a shape with circular cross-sections. At a height $h$ above the bowl's base, this cross-section has squared radius $25-(5-h)^2=10h-h^2$. You can can use the square root of this as your upper bound for $\rho$, but it's easier to use a single Integral. In cubic centimetres, the volume of water is $\int_0^2\pi(10h-h^2)=\frac{52\pi}{3}.$This is a believable proportion of the bowl's capacity, namely $\frac{26}{125}$.

Answer 3

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 $$ \int_0^{2\pi} \int_{\arccos(-3/5)}^\pi \int_{-3/\cos\phi}^5 \rho^2\sin\phi \,d\rho\,d\phi\,d\theta$$

Answer 4

Assume you are in anti-gravity space and just turn the bowl. Things will get easy by looking this way. Your $\theta$ bounds are correct. For $\phi$ bounds, you will start from 0 and will end up on $cos^{-1}(3/5)$ or $sin^{-1}(4/5)$.Try some geometry, it's easy. Now for $\rho$ , clearly upper bound is 5. To find lower bound, draw a ray from origin such that it intersects line y=3 on your paper.Value of $\rho$ depends on $\phi$. Again do some geometry to conclude lower bound is $3/cos\phi$ or simply $3sec\phi$.