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A right circular cylinder is at an incline of 15° from the horizontal and the liquid is level with the lowest point of the top rim of the can. The radius is 3.2004 cm and the height is 11.938 cm. What is the volume of the liquid?

I believe I should use integration with cross-sections of rectangles. The width of each rectangle would be the diameter of the cylinder. I believe my limits of integration would be from 0 to 4.7$\cdot\sin(15°)$ or 1.2164. I'm not sure how to figure out the changing lengths of the rectangles.

Am I on the right track?

$$\int_{0}^{4.7\sin(15°)}6.4008?dy$$

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  • $\begingroup$ My first thought is that the volume of liquid is the volume of the cylinder minus the volume of that portion of the cylinder lying above the lowest point of the top rim. $\endgroup$ – saulspatz Apr 30 '18 at 16:42
  • $\begingroup$ That makes sense as well. However, I still am lost as to how to find the formula. Any ideas? $\endgroup$ – user147485 Apr 30 '18 at 16:52
  • $\begingroup$ Where are you up to in your studies? Have you done multiple integrals? (Frankly, it's been longer than you can imagine since I've done one of these problems, and I'm not sure I still can, but I'd like to know what tools I can use.) $\endgroup$ – saulspatz Apr 30 '18 at 16:58
  • $\begingroup$ This is a calculus two problem and the section we are studying is finding the volume of solids by known cross-sections, disk method, or washer method (chapter seven). Multiple integration isn't introduced in our textbook until chapter fourteen. $\endgroup$ – user147485 Apr 30 '18 at 16:59
  • $\begingroup$ Then my first idea is not applicable. I was thinking of a double integral, which as you say, you will not study until chapter 14. I'll have to think about it for a bit. $\endgroup$ – saulspatz Apr 30 '18 at 17:07
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If you rotate the cylinder so the base is on the x axis and the liquid level is at 15 deg (positive slope), then the height of the rectangles is given by the height of the cylinder minus the evaluation of the 15 deg slope of the level surface (linear equation). Limits of integration can be x = 0, to x = d (cylinder diameter).The width of the rectangle will be the chord length of a line cutting the circumference of the circular cross section as it moves from left to right.

Using this I get$$\int_0^{6.4008} (.20795x+10.22291)(2\sqrt{10.24256-(3.2004-x)^2})dx$$ $= 350.3675 cm^3$

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Here is way how to get the volume of the empty part with a single integral. You will need

  • the area of an ellipse with semi-major and semi-minor axes $a$ and $b$: $$A = \pi a b$$

The water surface corresponds to a cut of a plane with the cylinder and is, hence, an ellipse. Now, the volume above water level can be modelled using ellipses. Just imagine that the angle between the plane and the top of the cylinder decreases.

Let $\nu_0$ be the angle between the water surface and the top of the cylinder.

Now, imagine for each $\nu \in [0,\nu_0]$ a plane cuts the cylinder as described above. You sum up the areas of all the ellipses created by the cuts: $$A_{\nu}= \pi a_{\nu} b_{\nu}$$ If $r$ is the radius of the cylinder, the main axes are $$b_{\nu} =r \mbox{ and } a_{\nu} = \frac{r}{\cos {\nu}}$$

Now, you get an infinitesimal volume slice (like an elliptic wedge) with respect to $\nu$ as folllows:

  • the infinitesimal height at the end of the slice is $\frac{2r}{\cos \nu} d\nu$ (at the end of the major axis)
  • the vertical section of an elliptic wedge is an infinitesimal right triangle (similar to the calculation of area in polar coordinates), which gives rise to a factor of $\frac{1}{2} $$$dV_{\nu} = A_{\nu} \frac{1}{2} \frac{2r}{\cos \nu} d\nu = \frac{\pi r^3}{\cos^2 \nu}d\nu$$

So, the volume of the empty part is

$$V = \int_0^{\nu_0}dV_{\nu}=\pi r^3\int_0^{\nu_0}\frac{1}{\cos^2 \nu}d\nu = \pi r^3 \tan \nu_0$$

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No calculus is needed––trig will do. I always hated when they posed simple problems and asked for a calculus answer.We need to find the height of the center of the liquid surface. Given this, you can use the standard cylinder volume formula to find the answer. To do this we must multiply the distance-to-center across the surface times the sine of 15 degrees and subtract it from the cylinder height. The radius of the cylinder divided by the liquid surface distance-to-center is the cosine of the angle.

The cosine of 15 degrees is (sqrt(6)+sqrt(2))/4 = 0.965925826. The distance across the surface to the center (the hypotenuse) is the radius divided by the cosine so 3.2004/0.965925826=3.313297889.

Then we find the sine of 15 degrees: (sqrt(6)-sqrt(2))/4=0.258819045. The surface radius times the sine is 3.313297889*0.258819045=0.857544595.

If you subtract this from the cylinder height you have the mean height of the liquid for your volume calculation. If the professor complains, I would ask for a problem that cannot be done without calculus.

The height at the center of the liquid is is 11.938-0.857544595= 11.08045541

The formula for volume of a cylinder is:

πr^2 h = 3.141592654*3.2004^2 *11.08045541= 356.5463596

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