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I wish to calculate the volume of this special cone ! The base is an ellipse ellipse of semi-major a and semi-minor b

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Thanks !

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    Integrate over the cross-sections. Assuming that all of the cross-sections are also ellipses of the same eccentricity, the area of each cross-section will be proportional to $y^2$. – 2017-02-21
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    You’re scaling uniformly, so each dimension of a cross-section is proportional to $y$. – 2017-02-24
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    I did this and now I have this integral to calculate (which is not simple!). I'm not sure if the result is correct: $$V = \frac{{4ab}}{\mu }\left( {c{e^{\mu a}}} \right)\mathop \smallint \limits_c^{c{e^{\mu a}}} ln\frac{z}{c}\;\sqrt {{a^2} - \frac{1}{{{\mu ^2}}}{{\left( {ln\frac{z}{c}} \right)}^2}} \;dz$$ http://math.stackexchange.com/questions/2157751/how-to-calculate-this-logarithm-integral?noredirect=1#comment4438917_2157751 – 2017-02-24
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    Sorry, I didn’t look carefully at the diagram and assumed that the $z$-axis was the cone’s axis. The area of each cross section is proportional to $(x/a)^2$ in this problem. Even so, the integral you came up with looks way too complex. – 2017-02-24

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There are a couple of ways to attack this sort of problem. Based on the diagram, we will assume $c>0$ and $\mu>0$.

Because the object is described as a “cone,” we will assume that the horizontal cross-sections are all homothetic ellipses, so one approach is to integrate these cross-sections over the height of the cone. The area of the base is $\pi a b$, so the area of each horizontal slice is some multiple of this. The semi-major axis is simply the $x$-coordinate of the slice which ranges from $0$ to $a$, so the required scale factor is $x/a$. We’re scaling uniformly, so this is also the scale factor for the semi-minor axis, hence the area of each slice is $\pi ab\left(\frac xa\right)^2$. We have $x=\frac1\mu\ln{\frac yc}$, so the volume is given by the integral $$V=\int_c^{c\exp(\mu a)}\pi ab\left(\frac1{\mu a}\ln{\frac yc}\right)^2\,dy = {\pi b\over\mu^2a}\int_c^{c\exp(\mu a)}\ln^2{\frac yc}\,dy.$$ This integral can be evaluated by making the substitution $u=y/c$ and then integrating by parts.

Another approach is to integrate over concentric cylindrical slices. We can find the volume of the region below the cone and subtract that from the volume of the enclosing elliptical cylinder, $\pi abc(e^{\mu a}-1)$. The lateral area of each cylindrical slice is $C(x)c(e^{\mu x}-1)$, where $C(x)$ is the circumference of an ellipse with major semi-axis length $x$ that is homothetic to the base of the cone. Unfortunately, there’s no closed-form expression for $C(x)$, but we have a way around that. We can find the volume using circular cross-sections and then stretch the coordinate axes. The lateral area of a cylindrical slice is simply $2\pi xc(e^{\mu x}-1)$. The major axis ($x$-axis in the illustration) needs no adjustment, but the minor axis needs to be scaled by a factor of $b/a$ to get our ellipse. So, we have $$V'=\frac ba\int_0^a 2\pi xc(e^{\mu x}-1)\,dx = {2\pi bc\over a}\int_0^a x(e^{\mu x}-1)\,dx$$ and finally $$V=\pi abc(e^{\mu a}-1)-{2\pi bc\over a}\int_0^a x(e^{\mu x}-1)\,dx.$$

I’ll leave evaluating one or both of these integrals to you.

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    Thank you very much Amd for your help, I'll try to evaluate the two integrales. But I didn't get this : so the area of each horizontal slice is $$\pi ab{(\frac{x}{a})^2}$$ could you explain this point please ? – 2017-02-27
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    @VolumeShape See my edit. – 2017-02-27
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    Thank you amd. This is the trick I had not found and so I arrived at an incalculable integral (I still have a little trouble with the $\pi ab{(\frac{x}{a})^2};$).This is the evaluation of the first integral : $$V = \frac{{\pi bc}}{{{\mu ^2}a}}\;\left[ {{\mu ^2}{a^2}{e^{\mu a}} - 2{e^{\mu a}}\left( {\mu a - 1} \right) - 2} \right]$$ But there's something wrong with the second one I did not get the same result ! I think the first approach is the correct one. – 2017-03-01
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    FWIW, I get the same answer with both. – 2017-03-02
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    Thank you very much Amd, it really helped me a lot – 2017-03-02
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    @ Amd: one more question please, if the axis of rotation is not x=0 but x = a, the area of slice would be then : $\pi ab{(\frac{x-a}{a})^2}$instead of $\pi ab{(\frac{x}{a})^2}$ is it correct ? – 2017-03-03
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    @ Amd: one more question please, if the axis of rotation is not x=0 but x = a, the area of slice would be then : πab(x−aa)2 instead of πab(xa)2 is it correct ? – – 2017-03-10
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    @VolumeShape This is not a surface/volume of revolution, so there’s no axis of rotation in the first place. If the center of symmetry is somewhere other than the origin, then you might translate it as it appears you’re trying to do. – 2017-03-13
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    Yes you're right I'm trying to translate it to x =a, so is it correct what i write then? $$\pi ab{(\frac{x-a}{a})^2}$$ – 2017-03-13
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    @VolumeShape As long as you also have $y=c\exp(\mu(x-a))$. – 2017-03-13
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    Why? This means when x = a, y = c !!! But in reality when x = a $y=c\exp(\mu(a)) $and when x=0, y = c. – 2017-03-13
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    finaly it look like this http://imgur.com/a/mB8wM thank you amd for your help – 2017-03-13
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    @VolumeShape That’s not a simple translation. Each horizontal slice will still have an area of the form $\pi a b p^2$, but you’re going to have to work out what $p$ is based on the maximum and minimum diameters as $y$ varies. – 2017-03-13
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    yes i get this, x varies from a to 2a, so y varies from f (a) to f (2a). The problem I can not find the area of each slice depending only on x or y and then do integral :(((((( !!!! – 2017-03-13
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    if we assume that the base is an circle (instead of an ellipse) and the revolution axe is x = R then : $$ v =\pi \;\mathop \smallint \limits_{f\left( R \right)}^{f\left( {2R} \right)} \ {\left( {x - R} \right)^2}dy$$ $$v =\pi \;\mathop \smallint \limits_{f\left( R \right)}^{f\left( {2R} \right)} {\left( {\frac{1}{\mu }ln\frac{y}{c} - R} \right)^2}dy$$ isnt it ? – 2017-03-13
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    up ! any help ? – 2017-03-22
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    Post a new question if you can’t work this new problem out for yourself. – 2017-03-23