# Math Help - Volume of a cone

1. ## 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.

2. 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: $V=\sqrt{2} \int_0^{2 \pi} \int_0^1 r(1-r) \ dr d \theta = \frac{\pi \sqrt{2}}{3}.$