- #1
brotherbobby
- 749
- 169
- Homework Statement
- If a liquid fills a hemispherical bowl of radius ##a## to a height ##h##, show that the volume of the liquid is ##\boxed{V = \dfrac{1}{3}\pi h^2(3a-h)}##. (See figure below).
- Relevant Equations
- 1. The equation of a sphere of radius ##a## : ##x^2+y^2+z^2=a^2##.
2. Volume is given by the triple integral : ##\displaystyle{V = \int\int\int dx \,dy \,dz}##
3. The volume of a cylinder of radius ##r## and height ##h## is : ##V = \pi r^2 h##
Drawing : The drawing of the problem is shown alongside.
This is also known as the problem of the "spherical cap". I have been able to do it by the normal way. It involves taking a thin slice of the liquid of thickness ##dz## at a height (or depth) ##z## from O and calculating its area ##A = \pi r^2(z)##, where ##r^2(z) = a^2-(a-z)^2##. Integrating this gives the volume : ##\displaystyle{V = \int\limits_{a-h}^h A\,dz}## and it gave the answer above in ##\boxed{\text{box}}##.
The issue : How to do the problem using the method of triple integrals ?
I draw the situation of the problem alongside, naming some points and flipping the diagram horizontally for convenience. The volume of the water is the spherical cap. This can be found by subtracting the volume of the cylinder ABDC from the volume of the dome APBDC : ##V_{\text{liquid}}\text{(APB)} = V_{\text{dome}}\text{(APBDC)} - V_{\text{cylinder}}\text{(ABDC)}##.
The radius of the liquid surface at the point ##h## is ##r = \sqrt{a^2-(a-h)^2}=\sqrt{2ah-h^2}\qquad {\color{blue}{(1)}}##
(I) Volume of the dome APBDC : Using triple intgrals,
##
\begin{align*}
V_{\text{dome}}\text{(APBDC)}&= \int\limits_0^a dz\hspace{-1em}\int\limits_{-\sqrt{r^2-z^2}}^{\sqrt{r^2-z^2}}dy\hspace{-1em}\int\limits_{-\sqrt{r^2-z^2-y^2}}^{\sqrt{r^2-z^2-y^2}}\hspace{-1em}dx\quad\small{\text{(Note the use of}\, r\,\text{the radius of the cap at}\, h)}\\
&= 4\int\limits_0^a \!dz\int\limits_0^r' \!\!dy\int\limits_0^{\sqrt{r'^2-y^2}}\hspace{-1em}dx\qquad(\text{where}\;r'^2=r^2-z^2)\\
&=4\int\limits_0^a dz\int\limits_0^r' dy{\sqrt{r'^2-y^2}}\\
&=\pi\int\limits_0^a dz(r^2-z^2)\qquad\qquad{\small{(\text{Upon using the known integral for} \int\sqrt{a^2-x^2}dx)}}\\
&=\pi\left|r^2z-\dfrac{z^3}{3}\right|_0^a\\
V_{\text{dome}}\text{(APBDC)}&= \pi\left[(2ah-h^2)a-\dfrac{a^3}{3}\right]\qquad{\color{blue}{(2)}}\\
\end{align*}
##
where I used the value of ##r## from ##(1)## above.
(II) Volume of the cylinder ABDC : The volume of the cylinder can be found without the need for integrating. We have
##
\begin{align*}
V_{\text{cylinder}}\text{(ABDC)}&= \pi r^2\times\text{height}\\
&= \pi(2ah-h^2)(a-h)\qquad\text{Using the value of}\, r\,\text{from}\, (1)\,\text{above}\\
V_{\text{cylinder}}\text{(ABDC)}&= \pi(2a^2h-ah^2-2ah^2+h^3)\qquad\color{blue}{(3)}\\
\end{align*}
##
Subtracting (3) from (2) above, we get the volume of the liquid or the "spherical cap" :
##\underline{V_{\text{liquid}}\text{(APB)} = \pi\left(2ah^2-h^3-\dfrac{a^3}{3}\right)}\quad{\color{red}{\huge\times}}##.
This does not match the answer : ##\boxed{V = \dfrac{1}{3}\pi h^2(3a-h)}\quad{\color{green}{\Large\checkmark}}##.
Request : Where did I go wrong in my calculation above?
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