Question

A ball of radius 13 has a round hole of radius 9 drilled through its center.

Answer 1

Answer:

Step-by-step explanation:

Given

radius of sphere $R=13\ units$

Hole is drilled with radius $r=9\ units$

Now this drilled hole will be in the form of Cylinder with height equals to diameter of sphere thus

Volume of drilled hole is

$V_d=\pi \cdot r^2\cdot h$

$V_d=\pi \cdot (9)^2\cdot 26$

$V_d=2106\pi$

Volume of Sphere

$V_s=\frac{4}{3}\pi R^3$

$V_s=\frac{4}{3}\pi (13)^3$

$V_s=2929.33\pi$

Remaining volume $V_{remaining}=2929.33\pi -2106\pi =823.33\pi$

$V_{remaining}=2586.90\ units^3$