how long to fill a propane tank to 1/4 full using volume of a solid of rotation

Question

I am having a huge problem with this homework problem. I have tried it a couple different ways but nothing I get is anything like a right answer. heres the problem,

 

A propane tank is in the shape generated by revolving the region enclosed by the right half of the graph of
x^2+16y^2=144 and the y-axis
about the y-axis. If x and y are measured in meters find the depth of the propane in the tank when it is filled to one-quarter of the tank's volume. (Round to 3 decimal places).

 

can anyone show me how to set this problem up and where to start integrating this problem and how to factor in the 1/4 full part?

 

Answer 1

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Answer 2

Noah,

I hope you've already found your answer. But if not, try this.

 

Note that the equation is an ellipse with standard equation x²/144 + y²/9 = 1

This ellipse has a horizontal major axis of 24 and a vertical minor axis of 6 with the center at the origin (0,0).

 

Using the right half tells us to use the positive x portion. When we rotate it around the y-axis, we get a series of concentric circles, all centered on the y-axis, stepping vertically from y = -3 to y = +3. So to find the volume, we need to add up the areas of these circles whose radii, at any given y value, will always be equal to the corresponding x value of the ellipse. Since the area of a circle is A = πr², we can state the areas for this scenario as:

 

A = πx²    

 

But since we fill a tank from the bottom to the top, we need to express this area in terms of y. So we  use the original eliptical equation to relate x and y as:

 

x² = -16y² + 144

 

Now we can restate the area equation as:

 

A = π(-16y² + 144)

 

and the volume will be the sum of those areas as y goes from -3 to positive 3.

           3

V =  ∫ π(-16y² + 144)          We can take the π outside of the integral.

         -3

 

             3

V = π∫ -16y² + 144

            -3

 

So when you integrate this part and solve, it is simpler to leave your answer in terms of pi. 1/4 of the volume will of course be the previous answer divided by four.

 

Now resolve the same integral from -3 to y₁ where is the y coordinate for 1/4 the volume, and you've substituted the calculated V/4 in for V. Just solve for y₁.  Being a cubic equation you'll find three zeroes, all unique and real in this case. Just remember that y₁ is not the depth.

 

d = y₁ - (-3)

 

If you think about the shape of your ellipse, it's easy to see that V/4 will occur somewhere between 1/4 of the depth and 1/2 of the depth, i.e. y₁ will NOT be a positive number. In any event, only one of the zeroes will fall with the range of the ellipse.

 

Let me know if you still have trouble.