A ball of radius 14 has a round hole of radius 4 drilled through its center. Find the volume of the resulting solid.
I keep getting 10086.6068 but its not right any help would be nice thanks
I read it as a round whole not a spherical hole. So think a drillbit of radius 4 straight through the center of the sphere. The sphere has a volume of $\displaystyle \frac{10976\pi}{3}$.
The cylindrical portion of the hole has a radius of 4 but what is the height? Draw a circle of radius 14 and inscribe in it a rectangle whose top and bottom have length 8 (a crossection of the sphere through a great circle "parallel" to the hole). Draw a radius to a point on the rectangle and and the circle (the corner of the rectanlge), and draw a vertical radius. Now you have a right trangle whose hypotenuse is length 14 and whose base is length 4.
Therefore the height of the cylinder is $\displaystyle h = 2\sqrt{180}$.
So the volume of this cylinder is $\displaystyle V_c = 32 \pi \sqrt{180}$
What tools do you have for calculating volume. Do you have integral calculus?
The solid you are looking for is a symmetric spherical layer with the central cylinder missing.
The volume of the layer is:
$\displaystyle V_{layer}=\frac{\pi h}{6} \cdot \left(6r^2+h^2\right)$
The volume of the central cylinder is
$\displaystyle V_{cyl}=\pi r^2 \cdot h$
Thus the volume of the solid is:
$\displaystyle V_{solid}=\frac{\pi h}{6} \cdot \left(6r^2+h^2\right) - \pi r^2 \cdot h = \frac16 \pi h^3$
$\displaystyle r = 2$ and I've got $\displaystyle h = 6 \sqrt{5}$
Hello, waite3
I must assume that this is a Calculus problem.
A ball of radius 14 has a round hole of radius 4 drilled through its center.
Find the volume of the resulting solid.Code:14| * * * *:::::|:::::* *:::::::|:::::::* *--------+--------* 4| * | * - - * - - - - + - - - - * - - * | * | *- - - - + - - - -* * | * * | * * * * |
We have a circle: .$\displaystyle x^2+y^2 \:=\:14^2\quad\Rightarrow\quad y^2 \:=\:196 - x^2$
We have the region bounded by the upper semicircle and $\displaystyle y = 4$
Revolve it about the x-axis to generate the desired solid.
The circle and horizontal line intersect at: $\displaystyle x \:=\:\pm6\sqrt{5}$
The volume of the solid is: .$\displaystyle V \;=\;2 \times \pi\int^{6\sqrt{5}}_0\bigg[(196 - x^2) - 4^2\bigg]\,dx$
And we have: . $\displaystyle V \;=\;2\pi\int^{6\sqrt{5}}_0\left(180-x^2\right)\,dx $
I get: .$\displaystyle 1440\pi\sqrt{5} \:\approx\:10,115.73 $