# Volume of a cone

• Jul 4th 2009, 04:16 PM
arbolis
Volume of a cone
I'm having an extremely hard time trying to calculate the volume of the cone given by $(z-1)^2=\frac{x^2}{2}+y^2, 0\leq z \leq 1$.
I've tried to draw it in the $xyz$ plane : its generators cross when $z=1$, $x=y=0$. In the $xy$ plane its cross section is an ellipse centered at the origin, whose semi major axis is worth $1$ unit and its semi minor axis is worth $1/2$. But I'm not sure I've done it right. Also I have a big problem : when using cylindrical coordinates, $\theta$ goes from $0$ to $2\pi$, $z$ goes from $0$ to $1$ but $r$ goes from $0$ to ... I'm unable to find it. I guess I must find some equation of generators, but still, the cone is not circular (cross section with the xy plane), making it difficult for me to find its volume.
• Jul 4th 2009, 04:50 PM
NonCommAlg
Quote:

Originally Posted by arbolis
I'm having an extremely hard time trying to calculate the volume of the cone given by $(z-1)^2=\frac{x^2}{2}+y^2, 0\leq z \leq 1$.
I've tried to draw it in the $xyz$ plane : its generators cross when $z=1$, $x=y=0$. In the $xy$ plane its cross section is an ellipse centered at the origin, whose semi major axis is worth $1$ unit and its semi minor axis is worth $1/2$. But I'm not sure I've done it right. Also I have a big problem : when using cylindrical coordinates, $\theta$ goes from $0$ to $2\pi$, $z$ goes from $0$ to $1$ but $r$ goes from $0$ to ... I'm unable to find it. I guess I must find some equation of generators, but still, the cone is not circular (cross section with the xy plane), making it difficult for me to find its volume.

since $z \leq 1,$ we get $z=1-\sqrt{\frac{x^2}{2}+y^2}.$ the volume is $V=\int \int_D \left(1-\sqrt{\frac{x^2}{2}+y^2} \ \right) \ dA,$ where $D=\{(x,y) \in \mathbb{R}^2: \ \frac{x^2}{2}+y^2 \leq 1 \}.$ put $x=\sqrt{2}r \cos \theta, \ y=r \sin \theta, \ 0 \leq \theta \leq 2 \pi, \ 0 \leq r \leq 1.$

under this transformation, $D$ is transformed to the unit disc and the Jacobian is $\sqrt{2}r.$ thus: $V=\sqrt{2} \int_0^{2 \pi} \int_0^1 r(1-r) \ dr d \theta = \frac{\pi \sqrt{2}}{3}.$