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.
Best Answer
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...?)