[Math] How to find the volume of oblique cone

calculusintegration

It is probably dublicated but couldnt find.
Let radius=r and height=h, find the volume of oblique cone with use of integral

if it was right cone I'd use $y=h-\frac{xh}{r}$ so $\pi\int_0^h(\frac{r(h-y)}{h})^2dy$.

is it same with oblique?

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The transformation which maps a right cone onto a oblique cone is described by the shear matrix which looks like so (k parameter determines the obliques):

$$ \left(\matrix{1\;0\;k\\0\;1\;0\\0\;0\;1}\right)$$

It has the determinant equal to $1$ therefore the formula for the cone's volume doesn't change. It remains

$$ V= \frac{hr^2\pi}{3} $$

Related Solutions

[Math] Finding volume of a cone through integration

Place your cylinder so the center of the base is at $(0,0)$, and the apex is at $(0,5)$.

If you imagine looking at the cylinder straight on, it will look like a triangle with base $20$ and height $5$. If you make a horizontal slice at level $y$, then you get a figure with two similar triangles:

                   ^                  ^
                  / \                 | 
                 /   \                |
    ^           /_____\               5 
    |          /   2r  \              |
    y         /         \             |
    |        /           \            |
    V       /_____________\           V

            |----- 20 -----|

then using similar triangles note that the height of the triangle on the top is $5-y$, and the base is $2r$. So we have $$\frac{2r}{20} = \frac{5-y}{5}.$$ From this, we can express $r$ in terms of $y$.

This is where you are using the fact that the base of your original cone is $10$ cm.

[Math] Deriving the Volume of a Cone using Solids of Revolution

I'm not sure about your setup, but in any case you should be integrating in a direction perpendicular to the $r$ in your cross section.

I'll use a setup that's hopefully what you had in mind: the tip of the cone is at the origin, and the cone is lying on its side. Suppose the radius of the base is $r$ (this is the $y$ value at the end of the triangle you're rotating) and the height is $h$ (this is the $x$ value at the end of the triangle).

Then, consider a circular cross section at some point $x$. The radius is the corresponding $y$ value, which by similar triangles is $\frac{rx}{h}$. So, the area is $$ A(x) = \frac{\pi r^2}{h^2} x^2, $$ and the volume is $$ V = \int_0^h \frac{\pi r^2}{h^2} x^2\ dx = \frac{\pi r^2 h^3}{3h^2} = \frac{1}{3} \pi r^2h. $$ Note the cone lies on its side, so the $x$ values we integrate over range from $0$ to the "height" of the cone, $h$.

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Related Solutions

[Math] Finding volume of a cone through integration

[Math] Deriving the Volume of a Cone using Solids of Revolution

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