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If i have frustum and its top is cut by an inclined plane at angle $\alpha$, such that it makes an ellipse. The height is $h$ (at the axis of obliquely truncated frustum). How can i use triple integral to determine its volume, and coordinates of its geometric center. I will be thankful. I can determine the volume of elliptical cone by using parametric equation however i am confused to obtain parametric equations for right circular cone cut by inclined plane. The radius of the bottom surface is $R$ and top surface is $r$, as shown figure below. Frustum Figure

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    In reply to your (deleted) post over at [volume of a truncated cone that is not a frustum](http://math.stackexchange.com/questions/2072846), the volume formula was found using geometry, not calculus. ;) (The solid extending up to the vertex is a cone with elliptical base and known height.) – 2017-01-24

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The equation of the conical surface can be written in cylindrical coordinates $(\rho, \theta, z)$ as $$ z={H\over R-r}(R-\rho), $$ whereas the equation of the plane is $$ z=(\tan\alpha) \rho\cos\theta+h. $$ By combining these one can find the equation of the projection of their intersection on the $xy$-plane: $$ \rho={R-(h/H)(R-r)\over 1+[(\tan\alpha)(R-r)/H]\cos\theta}, $$ which is the polar equation of an ellipse having a focus at $(0,0)$.

In your volume integration, you must integrate along $z$ from $0$ to the value given by the conical surface equation, if $(x,y)$ is outside the ellipse, and from $0$ to the value given by the plane equation, if $(x,y)$ is inside the ellipse.

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    Hello, can you please explain what do you mean when you said, "if (x,y) is outside ellipse and (x,y,) inside the ellipse". I did not understand this point. Also, if i am getting it right, then the limits for triple integral will be like: 0 to 2pi, 0 to rho, 0 to z. (sorry, i am having trouble in writing the equations format properly). Please explain a little @Aretino – 2017-01-26
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    Integration on $\rho$ and $z$ must be split into two parts: for $0\le\rho\le\bar\rho$ (where $\bar\rho$ is given by my third equation above) you are inside the ellipse and $0\le z\le z_1$, where $z_1$ is given by the second equation above; for $\bar\rho\le\rho\le R$ you are outside the ellipse and $0\le z\le z_2$, where $z_2$ is given by the first equation above. – 2017-01-26
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    Please correct me if there is a mistake i am making: $V$ = $V1$+$V2$ , where $V1$=$$\int_0^{2\pi}\int_\varrho^{R}\int_0^{z2} \varrho \,dz\,d\varrho\,d\vartheta$$ and $V2$ = $$\int_0^{2\pi}\int_0^\varrho\int_0^{z1} \varrho \,dz\,d\varrho\,d\vartheta$$. By $\varrho$ = rho over bar (given in third equation), $z2$ = first equation, $z1$ = second equation . Please correct me on this. @Aretino – 2017-01-26
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    It should be all right. – 2017-01-26
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    I am working on this way. I will ask for help if i face any problem. I also need to determine the center of the volume using this triple integral, would be possible to get center of volume this way? @Aretino – 2017-01-28
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    Of course: to find the coordinates of centroid the integrand must be $x=\rho\cos\theta$ or $z$. Notice that $y_{centroid}=0$ by symmetry. – 2017-01-28
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I think the easiest way to get the volume is to find the vertex of the cone from which the frustum is cut, find the volume of the cone between the vertex and the elliptical cut, and subtract that from the volume of the cone between the vertex and the farther face of the frustum.

But if you must have a triple integral, how about setting up a coordinate system better suited to the problem? Consider "conical" coordinates $(\lambda, \theta, z)$, defined in terms of Cartesian coordinates $(x,y,z)$ where the origin is at the vertex of the cone from which the frustum is cut, the $z$ axis is the axis of the cone, $\theta=\mathrm{atan2}(y,x)$, and $\lambda=\frac{\sqrt{x^2+y^2}}{z}.$ In these coordinates, the curved surface of the cone is just a surface with a constant value of $\lambda$. You can work out the unit of volume integration for this coordinate system; I get $\lambda z^2\,d\lambda\,d\theta\,dz.$

Find the equation of the inclined plane in this coordinate system in the form $z=f(\lambda, \theta)$, find $z=z_0$ at the bottom face of the frustum, and integrate $f(\lambda, \theta)-z_0$ over the values of $\lambda$ and $\theta$ on that face.

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    I wanted to work out using the procedure you suggested, but first thing i could not understand was how to get "Lambda = (\sqrt(x^2+y^2))/z$ ". can you please explain a little. then i will try to work on it. @David K – 2017-01-28
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    In your figure, orienting the $z$ axis downward, the face with radius $R$ has $z$ coordinate $z=RH/(R-r).$ The coordinates of points on the circumference of that face satisfy $\sqrt{x^2+y^2}=R,$ so at those points $\lambda=R/z=(R-r)/H.$ That is the maximum value of $\lambda$ anywhere in the cone: all points in the interior of the cone have smaller values of $\lambda.$ The minimum value is $0$ (on the $z$ axis), so the bounds of integration are $0$ to $(R-r)/H.$ – 2017-01-28
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    Incidentally, I think any triple integral is going to more difficult to set up and solve than the geometric solution given to your earlier question. So be prepared to work hard if you must do it that way. – 2017-01-28
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    I can also use the geometric solution if i can get a general equations for volume, and coordinates of center of volume. Because in m actual problem the inclination angle of the plane changes with time. – 2017-01-28
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    If I understand you now, in addition to the volume, you need to find the center of mass of a cone of uniform density that has been truncated at an angle. Rather than try to include that in the existing questions, I think at this time it would be better to post this as a new question. The question would be specifically how to find the center of mass of the object geometrically. You can link to the earlier questions to show that the volume and triple integration have already been asked (so you do not just get answers you already have). – 2017-01-28