Volume of Solid Enclosed by an Equation I'm having problems finding the triple integrals of equations. I guess it has to do with the geometry. Can someone solve the two questions below elaborately such that I can comprehend this triple integral thing once and for all:
Compute the volume of the solid enclosed by


*

*$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1, x=0,y=o, z=0$

*$x^2+y^2-2ax=0, z=0, x^2+y^2=z^2$
 A: Here's #2: the first equation is equivalent to
$$(x-a)^2+y^2=a^2,$$
which is a cylinder centered at $(a,0,0)$ parallel to the $z$-axis. It's intersection with the $xy$-plane is a circle; the volume we want is the solid above this circle and below $z^2=x^2+y^2$. Let the region bounded by this circle be $C$, so the volume is 
$$\iint_C \sqrt{x^2+y^2}\, dA=2\int_0^{\frac{\pi}{2}} \int_0^{2a\cos\theta}r^2 \, dr\, d\theta,$$
where we used polar coordinates, in which $r^2=x^2+y^2$ and $dA=r\,dr\,d\theta$ (loosely speaking). The bounds come from the fact that half of the region $C$ is also described by the polar curve $r=2a\cos\theta$, where $0\leq \theta\leq \frac{\pi}{2}$, and the multiplication by 2 accounts for the other half. Then, evaluating this, we get
$$2\int_0^{\frac{\pi}{2}} \left[\frac{r^3}{3}\right]_0^{2a\cos\theta}\, d\theta=\frac{16a^3}{3}\int_0^{\frac{\pi}{2}} \cos^3\theta\, d\theta.$$ 
This last integral can be evaluated using standard techniques, and turns out to be $\frac{2}{3}$ for a final answer of $\frac{32a^3}{9}$. 
A: for the #1 if you want to use a geometrical solution you can say that it's the volume of a pyramid with apex at $(0,0,0)$ intersecting the axes at $(a,0,0),(0,b,0),(0,0,c)$ now because of the orthogonality of axes we can say the base is a right triangle having the area $\frac{1}{2}ab$ and the length of the height is $c$ then:
$$volume\; of\; pyramid=\frac{1}{3}(area\; of\; base)\times height=\frac{1}{6}abc$$  

For a calculus solution we have an enclosed space limited by the planes $z=0$ and $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ from bottom and top so the range for $z$ will be $0\le z\le c-\frac{c}{a}x-\frac{c}{b}y$
consider the side of the pyramid that is lying on the $xy$-plane, it is a triangle enclosed by the lines $x=0$ and $y=0$ and $\frac{x}{a}+\frac{y}{b}=1$ on $xy$-plane, so the range for y will be $0\le y\le b-\frac{b}{a}x$
and finally the range for $x$ is $0\le x\le a$ so we have: 
$$\begin{align}
\int^{a}_{0}\int^{b-\frac{b}{a}x}_{0}\int^{c-\frac{c}{a}x-\frac{c}{b}y}_{0}dz\,dy\,dx&=\int^{a}_{0}\int^{b-\frac{b}{a}x}_{0}(c-\frac{c}{a}x-\frac{c}{b}y)dy\,dx\\
&=c\int^{a}_{0}[y-\frac{1}{a}xy-\frac{1}{2b}y^2]^{b-\frac{b}{a}x}_{0}dx\\
&=c\int^{a}_{0}(\frac{b}{2}-\frac{b}{a}x+\frac{b}{2a^2}x^2)dx\\
&=bc[\frac{1}{2}x-\frac{x^2}{2a}+\frac{x^3}{6a^2}]^{a}_{0}\\
&=bc(\frac{a}{2}-\frac{a}{2}+\frac{a}{6})\\
&=\frac{1}{6}abc
\end{align}$$
And about your question for #2 that you have commented in the other answer, I've created the following figure for you in matlab:

you should find the volume of the space enclosed by $z=0$ at bottom and by the cone $z^2=x^2+y^2$ at top within a cylinder along the z-axis which its base is a circle centered at $(a,0)$ and has radius $a$. The height of the cylinder will be $2a$  
Samir Khan has used the fact that this volume is the volume below the function $z=\sqrt{x^2+y^2}$ in the area of a circle centered at $(a,0)$ with radius $a$ then has used polar coordinates to solve the integral
