Volume and surface area of $3/4$ of a sphere

Question

Take for example a 3-D sphere cut horizontally into quarters:

How would I identify the volume and surface area of top $3$ horizontal cuts?

Would it just be $\frac34\cdot$volume of complete sphere and $\frac34\cdot$ surface are of complete sphere?

Thanks for your time.

Answer 1

No, the volume & total surface area of the sliced\cut sphere are not $3/4$ of the volume & surface area of the complete sphere but the curved surface area (excluding the area of the circular face) of the sliced sphere becomes $3/4$ of the surface area of the complete sphere.

In general, the volume $V$ & total surface area (including area of circular face) $A_t$ of frustum of sphere (i.e. sliced\cut by a plane) having a radius $R$ & vertical height $H$ (as shown in figure below) are given by

$$\boxed{\color{blue}{V=\frac{\pi}{3}(3R-H)H^2}}$$

$$\boxed{\color{blue}{A_t=\pi H(4R-H)}}$$ $$\boxed{\color{blue}{\text{Curved surface area, }A_s=2\pi RH}}$$ Where, $0\le H\le 2R$

As per your question, the vertical height of the sliced sphere is $H=\frac{3}{4}(2R)=\frac32R$. Plugging this value in the above-generalized formula, we get the volume & total surface area of the sliced sphere as follows $$V=\frac{\pi}{3}\left(3R-\frac32R\right)\left(\frac{3R}{2}\right)^2=\frac{9}{8}\pi R^3=\frac{27}{32} (\text{Volume of complete sphere})$$ $$A_t=\pi \cdot \frac32R\left(4R-\frac32R\right)=\frac{15}{4}\pi R^2=\frac{15}{16}(\text{Surface area of complete sphere})$$ While the curved surface area $V_s$ of the sliced sphere $$A_s=2\pi R\left(\frac32R\right)=3\pi R^2=\frac{3}{4}(\text{Surface area of complete sphere})$$

Answer 2

You could integrate.

$$\int_V dV=\int A(x)\,dx$$ where $A(x)$ is the area of a slice of the sphere. Since this is a circle, $A(x)=\pi r^2$ where $r$ is the radius of this slice.

What is the radius there? We can get it from the Pythagorean theorem: sphere slice

So we are computing $$\int_V dV=\int A(x)\,dx=\int \pi(\sqrt{R^2-x^2})^2\,dx=\int \pi(R^2-x^2)\,dx=\pi\int R^2-x^2\,dx$$

The bounds of the integral depend on which of the slices you're looking to find the volume of. In the case of the bottom slice in your picture, we want to go from $0$ to $R/2$.

$$ \begin{eqnarray*} \pi\int_0^{R/2}R^2-x^2\,dx&=&\pi\left(\left.R^2x-\frac{x^3}{3}\right|_0^{R/2}\right)\\ &=&\pi\left(\frac{R^3}{2}-\frac{R^3}{24}\right) \end{eqnarray*}$$

Modify the bounds and recalculate this to get the area of the first and second slices.

Can you modify this argument to compute the surface area? (Hint: the area of a each slice adds up to be the volume, so for the surface area, you want to use the ____ of each slice...?)

Answer 3

Area proportionality/linearity with water height is valid for Area but not Volume. Derivatives of functions typically illustrate this.

Imagine a spherical tank (constant $R$ ) being filled from bottom by means of a flexible garden hose. We monitor with respect to water column height representing sections you mentioned. We have for the wetted area

$$ A= 2 \pi R h$$ $$ \dfrac {dA}{dh} = 2 \pi R \tag1 $$ which is a constant, this is linear dependence. ( btw, Archimedes was the first to state this).

However,

$$ V= \dfrac{\pi h^2}{3}(3R-h) $$ $$ \dfrac {dV}{dh} = \pi h (2R-h) \tag2 $$

FractionalVolume $(=Volume/\frac43 \pi R^3)$is plotted against Fractional Area $(=Area/4 \pi R^2)$

In answer to your question, fractional areas and volumes have to be compared.

Half Volume is filled when half Area is wetted. When quarter Area wetted, less than quarter Volume is filled. But when $\frac34$ Area is wetted more than $\frac34$ Volume is filled.

The rates at which volume increases is neither constant nor proportional to height $h$. The graph above shows the volume growth rate is zero when sphere starts to fill and when full, but maximum when water level is on the equator.