## A cylindrical tank for for storing water has a vertical diam

Limits, differentiation, related rates, integration, trig integrals, etc.
stellar
Posts: 21
Joined: Wed Mar 25, 2009 7:40 pm

### A cylindrical tank for for storing water has a vertical diam

A cylindrical tank for for storing water has a vertical diameter of ten feet. What percentage of the total capacity is being used when the depth of the water in the tank is seven feet?

The whole volume is pi*r^2*L=25piL with L being the length that we don't know (and I don't think we need?).

But how do i find the volume for 7ft?

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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Draw the cross-sectional view of the tank, being a circle. The radius will be $r = 5$. You can easily find the area of the half-circle which is filled, being the five feet from the bottom to the halfway mark: find the area of the whole circle, and divide by 2. The hard part is the last two feet.

One method would be to draw your circle centered on the origin. Then the top half of the circle is given by:

. . . . .$y\, =\, \sqrt{25\, -\, x^2}$

By noting the intersection of this curve with the horizontal line $y\, =\, 2$, you can find the $x$-values for the intersections. Then integrate, as you find most useful. (I would integrate in the first quadrant, from $x\, =\, \sqrt{21}$ to $x\, =\, 5$, multiply by 2 to get the identical portion in the second quadrant, and then multiply to find the area of the rectangle in between these odd-sized ends.)

Once you have the entire cross-sectional area, divide this by the total area. (You don't need the length of the tank, since the value of the length would cancel out when you did this division.) The result is the percentage of the total that is currently in use.

Don't forget to convert the decimal answer to percentage form.