Find the volume of the shapes defined as: $n$-dimensional cone
Can this be done without using spherical coordinates?
$$i)\ \frac{x_1^2}{a_1^2}+\cdots+\frac{x_{n-1}^2}{a_{n-1}^2}=\frac{x_n^2}{a_n^2}, \quad 0\leq x_n \leq a_n.$$
$$ii)\ \text{The shape bounded by the following:} \\ \sqrt{\frac{x}{a}} + \sqrt{\frac{y}{b}}=1, \quad \sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=2, \\ \frac{x}{a} = \frac{y}{b}, \quad 4 \frac{x}{a} = \frac{y}{b} \  \ (a,b > 0)$$
Im thinking that some of these last equations are cones too? Im not sure. What geometric shapes are they to begin with? Does anyone have a link to a site, that explains these shapes?
 A: It can be, but it's usually not because the equations are really messy. Almost no problem is only solvable in a given coordinate system, it's just a question of which integrals are less awful
A: (i) If we take $n=3$, implicit equation 
$$\frac{x_1^2}{a_1^2}+\frac{x_{2}^2}{a_{2}^2}=\frac{x_3^2}{a_3^2},0\leq x_3 \leq a_3$$
describes a $H_1$, a (truncated) hyperboloid of one sheet. 
see for the volume computation (using Cavalieri's principle):
http://www.cut-the-knot.org/Generalization/Cavalieri5.shtml
Beyond 3D, generalized spherical coordinates, like in 4D:
$$\begin{cases}
x_1 & = & R\sin\psi\sin\phi\cos\theta \\
x_2 & = & R\sin\psi\sin\phi\sin\theta \\
x_3 & = & R\sin\psi\cos\phi \\
x_4 & = & R\cos\psi
\end{cases}
$$
can be adapted to such volumes. But I have no general answer.
ii) This is very different: it is the area delimited by two truncated parabolas  and two straight lines (it is not usual to speak about cones in 2D). You have to determine four points of intersection $A,B,C,D$ and then dissect areas under the curves... Here is a figure in the case $a=5$ and $b=8$.

