Volume of liquid in a tank using a double integral

[mathmath]

Homework Statement

The fluid level in the tank ((1/4)*(x-4)^2 + y^2 == 4) is 7 m on the left edge of the tank (where x=0) and 5 m on the right edge (where x=8). Find the equation of the plane of the liquid, and use a double integral to find the volume of liquid in the tank. [Hint: you should use a "dy dx" iterated integral, where the bounds on y depend on x, and are given by the equation of the base of the cylinder.]

Homework Equations

tank equation: (1/4)*(x-4)^2 + y^2 == 4

The Attempt at a Solution

I have found the equation of the plane of liquid to be: z=7 - 2 x/8, I am not sure if this is correct. I am confused on what the bounds would be for the double integral and what would be the integrand since the tank equation is set equal to a number.

Thanks!

 

[mighty2000]

Dear forum post author,

Thank you for sharing your problem with us. You are on the right track with your equation for the plane of liquid. To find the volume of liquid in the tank, we can use a double integral with the bounds of integration being the base of the tank and the top surface of the liquid.

To set up the double integral, we first need to express the tank equation in terms of y. We can rearrange the equation to get y = ±√(4 - (1/4)*(x-4)^2). This gives us the bounds of integration for y as ±√(4 - (1/4)*(x-4)^2). For the bounds of integration for x, we can use 0 and 8, since those are the left and right edges of the tank.

Now, for the integrand, we can use the equation of the plane of liquid that you found, z = 7 - 2x/8. This will give us the height of the liquid at any point (x,y) in the tank.

Putting it all together, we get the double integral:

∫∫[7 - 2x/8] dy dx

Where the bounds of integration are:

0 ≤ x ≤ 8
-√(4 - (1/4)*(x-4)^2) ≤ y ≤ √(4 - (1/4)*(x-4)^2)

Evaluating this integral will give us the volume of liquid in the tank.

I hope this helps. Good luck with your calculations!