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WHITEO · Dec 29, 2006, 07:38 AM

A cylindrical tank has been buried underground in a horizontal position. The length of the tank is 30 feet, and the diameter of the tank is 16 feet. The volume of the liquid in the tank is calculated by measuring the depth of the liquid in inches with a long stick. Write a general formula for the volume of the cylinder (answer in cubic feet) in terms of d, the depth of the liquid.

Sample Depths and Volumes

Depth (in.)Volume (cu. ft.)
9 in. 102.449 ft3
15 in. 218.291 ft3
51 in. 1284.211 ft3
60 in. 1610.429 ft3
96 in. 3015.929 ft3
117 in. 3849.181 ft3
132 in. 4421.429 ft3
141 in. 4747.647 ft3
164 in. 5487.221 ft3
192 in. 6031.858 ft3

galactus · Dec 29, 2006, 08:25 AM

I assume you're required to use calculus.

Center the end of the tank at the origin.

Use the equation of a circle to construct a integral.

The radius of the tank is 8 and the height is 30.

$60\int_{-8}^{d-8}\sqrt{64-x^{2}}dx$

Perform the integration and you have a formula for the volume in terms of d(the depth).

For the integration, trig substitution may be a good approach. Let $x=8sin({\theta}) \;\ and \;\ dx=8cos({\theta})$

$60\int\sqrt{64-64sin^{2}({\theta})}8cos({\theta})d{\theta}$

$60\int\sqrt{64(1-sin^{2}({\theta})}8cos({\theta})d{\theta}$

Remember, $1-sin^{\theta}=cos^{2}{\theta}$

$60\int\sqrt{64cos^{2}{\theta}}8cos{\theta}d{\theta }$

$3840\int{cos^{2}{\theta}}d{\theta}$

Don't forget to change the limits of integration:

$-8=8sin{\theta}\Rightarrow{{\theta}=\frac{-\pi}{2}$

$d-8=8sin{\theta}\Rightarrow{{\theta}=sin^{-1}(\frac{d-8}{8})$

So, you have:

$3840\int_{\frac{-\pi}{2}}^{sin^{-1}(\frac{d-8}{8})}cos^{2}{\theta}d{\theta}$

Now use the identity: $cos^{2}{\theta}=\frac{1}{2}(1+cos(2{\theta}))$

$1920\int_{\frac{-\pi}{2}}^{sin^{-1}(\frac{d-8}{8})}(1+cos(2{\theta})d{\theta}$

= $960(sin(2{\theta})+2{\theta})$

Now use your limits of integration to derive the frmula you need.