I missed two classes in calculus and we're on a subject that I do not understand at all. If someone could just walk me through this problem I could probably begin to comprehend the rest.

The base of a solid elliptical cylinder is given by $ (x/5)^2 + (y/3)^2 = 1.$ A solid is formed by cutting off or removing some material such that the cross-sections perpendicular to the x-axis are all squares. Find the volume of such a solid.

  • $\begingroup$ @ Nathan T. Edited hoping you meant the sections of a cylinder perpendicular to x-axis as rectangles with mid-section a square of side 10. $\endgroup$ – Narasimham Feb 22 '15 at 18:40
  • $\begingroup$ The problem just says that the cross-sections perpendicular to the x-axis are squares. $\endgroup$ – Nathan T. Feb 22 '15 at 18:43
  • $\begingroup$ It is clearly impossible, you might be missing something/s. $\endgroup$ – Narasimham Feb 22 '15 at 18:46
  • $\begingroup$ Yeah, it says the area is enclosed by x^2/25 + y^2/9 = 1 and that the cross sections perpendicular to the x-axis are squares. I am quite confused. $\endgroup$ – Nathan T. Feb 22 '15 at 18:50
  • $\begingroup$ You see it is not a circular cylinder but an elliptic one..Shall re-edit it. $\endgroup$ – Narasimham Feb 22 '15 at 19:10

First imagine how you'd do this problem in 2 dimensions. You need to set up a double integral over some x interval and then some y interval. The equation you're working with is in the form of an ellipse centered at the origin. Sections perpendicular to the x axis have constant x value, so you want x to be your last integral. So slice up your ellipse from left to right in segments of dx thickness from -5 to 5 and length $l$.

As you move from left to right, the length $l$ of the infinitesimal segments you are summing up changes. You can find this change just by looking at your equation. Solve for y and you get $y= \pm3\sqrt{1-\frac{x^2}{25}}$. So for some specific x value, these are the ranges y goes from. In two dimensions you would be able to set up your integral at this point.

In this case the integral is over 3 dimensions. but we're told each cross-section perpendicular to the x axis is a square. Therefore, if y goes from some -a to a, then z goes from the same exact -a to a. In other words, the bounds on z will be exactly the same as the bounds on y. This allows us to set up the integral.

$$ Volume=\int_{-5}^5\int_{-3\sqrt{1-\frac{x^2}{25}}}^{3\sqrt{1-\frac{x^2}{25}}}\int_{-3\sqrt{1-\frac{x^2}{25}}}^{3\sqrt{1-\frac{x^2}{25}}}dzdydx $$

First step is to integrate over z. $\int dz = z$, so the integral becomes

$$ Volume = \int_{-5}^5\int_{-3\sqrt{1-\frac{x^2}{25}}}^{3\sqrt{1-\frac{x^2}{25}}}dydx. $$

Now integrate over y. No y variables have been introduced into the equation yet, so this is also just a simple $\int dy=y$, and we have $$ Volume = \int^5_{-5} 4 * 9 (1-\frac{x^2}{25})dx .$$

The last integral is a piece of cake.


$$ y = 3 \sqrt{ 1 -(x/5)^2 } $$

$$ \int dA = 4 \int_{-5} ^5 9 ( 1 -(x/5)^2 ) dx $$

and you can take it from there.


To 3D visualize sections perpendicular to x-axis are squares: SquareXnSolid

  • $\begingroup$ Pretty sure you set this up backwards. Cross-sections perpendicular to the x-axis will have constant y values and varying x values, so the squares they form will have total area 2x*2x, not 2y*2y. $\endgroup$ – MonkeysUncle Feb 22 '15 at 19:31
  • $\begingroup$ @MonkeysUncle: Visualized as above. $\endgroup$ – Narasimham Feb 22 '15 at 20:20
  • $\begingroup$ Never mind me, I'm an idiot. $\endgroup$ – MonkeysUncle Feb 22 '15 at 20:23
  • $\begingroup$ How did you produce this figure? It is very helpful in understanding what the solid looks like. $\endgroup$ – user84413 Feb 22 '15 at 20:24
  • $\begingroup$ @MonkeysUncle.Never mind,it happens to us all. $\endgroup$ – Narasimham Feb 22 '15 at 20:32

You can use the fact that the volume can be found by integrating the areas of the cross-sections.

In this case, the base of the solid is the region bounded by an ellipse, and the cross-sections are squares with sides $s=2\left(\frac{3}{5}\right)\sqrt{25-x^2}$, so

$\displaystyle V=\int_{-5}^{5}A(x)\;dx=\int_{-5}^{5}\left(\frac{6}{5}\sqrt{25-x^2}\right)^2dx=\frac{36}{25}\int_{-5}^5(25-x^2)\;dx=\frac{72}{25}\int_0^5 (25-x^2)\;dx$.


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