# Math Help - drilled sphere

1. ## drilled sphere

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

2. Originally Posted by waite3
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
Since the volume of a sphere is (4/3)Πr^3, if you find the volume of the larger sphere [I get (10976/3)Π] and then subtract the volume of the smaller sphere [I get (256/3)Π], then you have found the volume of the resulting solid. [I get (10720/3)Π which is ≈ 11, 225.96]

3. Originally Posted by waite3
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
Volume(Sphere 1) = $\frac{10976\pi}{3}$

Volume (Sphere 2) = $\frac{256\pi}{3}$

Volume of Resulting Solid = $\frac{10720\pi}{3}$

4. Originally Posted by waite3
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
Originally Posted by TheHolly
Since the volume of a sphere is (4/3)Πr^3, if you find the volume of the larger sphere [I get (10976/3)Π] and then subtract the volume of the smaller sphere [I get (256/3)Π], then you have found the volume of the resulting solid. [I get (10720/3)Π which is ≈ 11, 225.96]
But the volume that has been removed isn't a sphere, is it? I would think it's a cylindrical solid with rounded end caps.

-Dan

5. ya the shape that is removed would be cylindrical with rounder caps but im not sure how to find the volume of that.

6. 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 $\frac{10976\pi}{3}$.

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 $h = 2\sqrt{180}$.

So the volume of this cylinder is $V_c = 32 \pi \sqrt{180}$

What tools do you have for calculating volume. Do you have integral calculus?

7. Originally Posted by waite3
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
The solid you are looking for is a symmetric spherical layer with the central cylinder missing.

The volume of the layer is:

$V_{layer}=\frac{\pi h}{6} \cdot \left(6r^2+h^2\right)$

The volume of the central cylinder is

$V_{cyl}=\pi r^2 \cdot h$

Thus the volume of the solid is:

$V_{solid}=\frac{\pi h}{6} \cdot \left(6r^2+h^2\right) - \pi r^2 \cdot h = \frac16 \pi h^3$

$r = 2$ and I've got $h = 6 \sqrt{5}$

8. 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: . $x^2+y^2 \:=\:14^2\quad\Rightarrow\quad y^2 \:=\:196 - x^2$

We have the region bounded by the upper semicircle and $y = 4$
Revolve it about the x-axis to generate the desired solid.

The circle and horizontal line intersect at: $x \:=\:\pm6\sqrt{5}$

The volume of the solid is: . $V \;=\;2 \times \pi\int^{6\sqrt{5}}_0\bigg[(196 - x^2) - 4^2\bigg]\,dx$

And we have: . $V \;=\;2\pi\int^{6\sqrt{5}}_0\left(180-x^2\right)\,dx$

I get: . $1440\pi\sqrt{5} \:\approx\:10,115.73$

9. nice thanks a lot the answer was 10,115.73