1. ## cylinder with water

A cylindrical water tank with radius two feet and length six feet is filled with water to a depth of 3 feet when in a horizontal position. If the tank is turned upright, what is the depth of the water? Give your answer in terms of pie.

Find the volume and total area of a cube with edge 2k.

A right triangular prism has height 20 and base edges 5, 12, and 13. Find the total area.

2. Why would someone spell $\pi$ as pie?. You can't really think it's spelled that way?.
It's not a pastry, it's a letter of the Greek alphabet. PI

There's different methods to find the solution to #1. Calculus or non-calculus.

The height of the empty region in the tank is given by $1=2(1-cos(\frac{\theta}{2}))$

Solving for theta, we find $\theta=2cos^{-1}(\frac{1}{2})=\frac{2\pi}{3}$

The volume of a circular segment is given by $\frac{1}{2}r^{2}(\theta-sin{\theta})=2(\theta-sin{\theta})=2(\frac{2\pi}{3}-sin(\frac{2\pi}{3}))=\frac{4\pi-3\sqrt{3}}{3}\approx{2.456....}$

The length of the tank is 6, so the volume of this portion of the tank is

$8\pi-6\sqrt{3}$

Subtract this from the volume of the tank which is $24\pi$

and get $24\pi-(8\pi-6\sqrt{3})\approx{60.66}$

Then use the volume of a cylinder formula and solve for h:

$4{\pi}h=24\pi-(8\pi-6\sqrt{3})$

$h\approx{4.83}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Calc method:

$12\int_{-2}^{1}\sqrt{4-x^{2}}dx\approx{60.66...}$

Now, the same method as before and we get the same answer.

3. Whoops, sorry... wasn't thinking. I have habit of that--I've spelled 'edition' as addition, etc...
Thank you!

4. Since I'm supposed to put my answer in terms of pi, can I simplify this statement further? Could I just divide everything by two?

$
4{\pi}h=24\pi-(8\pi-6\sqrt{3})
$

Oh, and does anybody want to try the other two problems?

5. Originally Posted by blindspot
...
Find the volume and total area of a cube with edge 2k.

A right triangular prism has height 20 and base edges 5, 12, and 13. Find the total area.
Hello,

as you may know the volume of a cube is calculated by:
$V_{cube}=a^3$. (a = length of edge). Plug in the given value and you'll get V = 8k³

The surface area of a cube is
$A_{cube}=6 \cdot a^2$. Plug in the given value and you'll get A = 24k²
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

The base triangle is a right triangle because 5² + 12² = 13². Thus the base area and the top area form a rectangle with length = 12 and width = 5.

The surface area of a right prism is calculated by:

$A_{prism}=base + top + perimeter_{base} \cdot height$. Plug in the given value and you'll get:

$A_{prism}=5 \cdot 12 + 30 \cdot 20 = 660$

6. Originally Posted by blindspot

Oh, and does anybody want to try the other two problems?
If you have other problems, feel free to post them in new threads.

-Dan