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I have a cut paraboloid made from parabola $y(x)=c+x-ax^2$ and $x = 0$ line. How do I compute volume of this cut paraboloid? I researched on Wolfram. see formula 16 and 17

yW0K5o
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2 Answers2

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It may help to take a step back and look at the basic definitions. The area and centroid are given by

$$ A=\int\!\!\!\int dy~dx=\int y(x)~dx\\ R_x=\frac{\int\!\!\!\int x~dy~dx}{\int\!\!\!\int dy~dx}=\frac{1}{A}\int x~y(x)~dx\\ R_y=\frac{\int\!\!\!\int y~dy~dx}{\int\!\!\!\int dy~dx}=\frac{1}{2A}\int y^2(x)~dx\\ $$

And finally, Pappus's $2^{nd}$ Centroid Theorem states that the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $R$, i.e., $V=2πRA$. Therefore, for rotation about the $y$-axis, we can say that

$$V=2\pi\int_0^{x_{max}} x~y(x)~dx$$

where $x_{max}$ is the point on positive $x$-axis where $y=0$, i.e.

$$x_{max}=\frac{1+\sqrt{1+4ac}}{2a}$$

Thus

$$V=2\pi\int_0^{x_{max}} x~(c+x-ax^2)~dx$$

You should be able to take it from here.

Cye Waldman
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  • Your solution and [the solution I accepted] https://math.stackexchange.com/a/2358978/356852 gives different results. Why? $\pi\int{[k+\sqrt{l-my}]^2}dy$ – yW0K5o Jul 27 '17 at 17:13
  • @AlexGawkins The desired result can be found equivalently by $$V=2\pi\int xy(x)~dx$$ and $$V=\pi \int x^2(y)~dy.$$ In your comment, I do not think you have correctly represented $x(y)$. My solution is correct; I have verified it numerically for random $a$ and $c$. You have accepted a solution that wasn't really a solution at all, was it? I recommend that you change that. – Cye Waldman Jul 27 '17 at 19:07
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This volume can be calculated using disk or shell method.

Paul's online notes is a good source, here is a link with many examples:

http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithRings.aspx

ThePortakal
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