# I need help setting up this problem! please!

• September 12th 2013, 11:54 AM
kyliealana
I need help setting up this problem! please!
A cylindrical water tank with radius 2m is installed in such a way that the axis is horizontal and the circular cross sections are vertical. Water is put into the tank so that the depth of the water is 3m. What percentage of the total capacity of the tank is being used? Round off to the nearest percentage point.
• September 12th 2013, 01:09 PM
HallsofIvy
Re: I need help setting up this problem! please!
Since the axis is horizontal, every horizontal cross section is a rectangle with length "H" and width depending on the height. To get the width, imagine the end as a circle with center at (0, 0) and radius 2: $x^2+ y^2= 4$ so that $y= \pm\sqrt{4- x^2}$. That is, each such horizontal cross section will have area $H\sqrt{4- x^2}$. To find the total volume integrate from -2 to 2. To find the depth of the water, integrate from -2 to 1.
• September 12th 2013, 07:42 PM
kyliealana
Re: I need help setting up this problem! please!
So I integrate ((4-x^2)^.5)dy? I am still slightly confused as to what I have to do.
• September 12th 2013, 11:13 PM
kyliealana
Re: I need help setting up this problem! please!
Attachment 29164

I feel like I am offtrack on the problem.
• September 13th 2013, 05:45 AM
HallsofIvy
Re: I need help setting up this problem! please!
The volume of the tank is $\int_{-2}^2 H\sqrt{4- x^2} dx$. The volume of water is $\int_{-2}^1 H\sqrt{4- x^2}dx$.
To find the percentage, divided the volume of water by the volume of the tank.