Answer:
Part 1) [tex]324\pi\ units^{2}[/tex] ------> [tex]7,776\pi\ units^{3}[/tex]
Part 2) [tex]36\pi\ units^{2}[/tex] ------> [tex]288\pi\ units^{3}[/tex]
Part 3) [tex]81\pi\ units^{2}[/tex] ------> [tex]972\pi\ units^{3}[/tex]
Part 4) [tex]144\pi\ units^{2}[/tex] ------> [tex]2,304\pi\ units^{3}[/tex]
Step-by-step explanation:
we know that
The largest cross sectional area of that sphere is equal to the area of a circle with the same radius of the sphere
Part 1) we have
[tex]A=324\pi\ units^{2}[/tex]
The area of the circle is equal to
[tex]A=\pi r^{2}[/tex]
so
[tex]324\pi=\pi r^{2}[/tex]
Solve for r
[tex]r^{2}=324[/tex]
[tex]r=18\ units[/tex]
Find the volume of the sphere
The volume of the sphere is
[tex]V=\frac{4}{3}\pi r^{3}[/tex]
For [tex]r=18\ units[/tex]
substitute
[tex]V=\frac{4}{3}\pi (18)^{3}[/tex]
[tex]V=7,776\pi\ units^{3}[/tex]
Part 2) we have
[tex]A=36\pi\ units^{2}[/tex]
The area of the circle is equal to
[tex]A=\pi r^{2}[/tex]
so
[tex]36\pi=\pi r^{2}[/tex]
Solve for r
[tex]r^{2}=36[/tex]
[tex]r=6\ units[/tex]
Find the volume of the sphere
The volume of the sphere is
[tex]V=\frac{4}{3}\pi r^{3}[/tex]
For [tex]r=6\ units[/tex]
substitute
[tex]V=\frac{4}{3}\pi (6)^{3}[/tex]
[tex]V=288\pi\ units^{3}[/tex]
Part 3) we have
[tex]A=81\pi\ units^{2}[/tex]
The area of the circle is equal to
[tex]A=\pi r^{2}[/tex]
so
[tex]81\pi=\pi r^{2}[/tex]
Solve for r
[tex]r^{2}=81[/tex]
[tex]r=9\ units[/tex]
Find the volume of the sphere
The volume of the sphere is
[tex]V=\frac{4}{3}\pi r^{3}[/tex]
For [tex]r=9\ units[/tex]
substitute
[tex]V=\frac{4}{3}\pi (9)^{3}[/tex]
[tex]V=972\pi\ units^{3}[/tex]
Part 4) we have
[tex]A=144\pi\ units^{2}[/tex]
The area of the circle is equal to
[tex]A=\pi r^{2}[/tex]
so
[tex]144\pi=\pi r^{2}[/tex]
Solve for r
[tex]r^{2}=144[/tex]
[tex]r=12\ units[/tex]
Find the volume of the sphere
The volume of the sphere is
[tex]V=\frac{4}{3}\pi r^{3}[/tex]
For [tex]r=12\ units[/tex]
substitute
[tex]V=\frac{4}{3}\pi (12)^{3}[/tex]
[tex]V=2,304\pi\ units^{3}[/tex]
