Questions About Integrals I've started moving into integral calculus and I am a bit confused about double and triple integrals. If double integrals are for finding the area of a 3 dimensional space between a graph what do triple integrals do?  Sorry if this question doesn't make much sense, I'm not always the best at expressing my questions.
 A: Among others, simple integrals can be used to


*

*find the length of a 2D or 3D curve,

*find the area between a 2D curve and a coordinate axis.


Double integrals can be used to


*

*find the area of a surface in 3D,

*find the volume between a surface and a coordinate plane in 3D.


Triple integrals can be used to 


*

*find the volume of a solid in 3D.


(I prefer not to mention "find the hypervolume between an hypersurface and a coordinate hyperplane".)
A: I believe you should change your perspective little.  Integration is essentially a summing process.  The real question is, "what are you summing? "
Double integrals can be used to find areas and volumes depending upon what you are integrating.
$\iint 1 dA $ can be thought of summing blocks height one and so directly corresponds to the area $dA$.
$\iint f(x, y) dA$ sums blocks of height $f(x, y) $ and therefore corresponds to volume.
Hope this helps. 
A: If you have a double integral, like this:
$\int \int f(x) dy dx$
...then this integral calculates a surface area. Typically, in a 3-dimensional space.
To get a better understand, you can imagine it as if the integrals would be sums: $\sum \sum f(x) \Delta x \Delta y$.
Or you can see it, if you have some programmer experience, if it would be a double loop to iterate over a surface in very small fragments, and summing the result.
Triple integrals, like this:
$\int \int \int f(x) dz dy dx$
do the same, and you can see them similarly, but here we have triple sums and triple loops.
The trick is with the integrals, that these fragments are infinitely small.
A: You should be aware of:
$$\int_{a}^{b}dx=b-a \hspace{190px} \text{interval length}$$ 
$$\int_{C}ds=\int_{a}^{b}\sqrt{1+y(t)'^2}dt \hspace{140px} \text{curve length}$$ 
$$\iint_{R}dA\hspace{290px}\text{region area}$$
$$\iint_{S}dS=\iint_{R}\sqrt{1+[f_x(x,y)]^2+[f_y(x,y)]^2}dA\hspace{10px}\text{surface area}$$
$$\iiint_{Q}dV\hspace{290px}\text{solid volume}$$
Now imagine you have $\int_{a}^{b}f(x)dx$, you are computing the sum of the inifinitesimal dx and multiplying that times f(x) (so the area under the curve provided f(x)>=0), if you take $\iint_{R}f(x,y)dA$ you multiply the area of each area element (each little rectangle in the plane xy) times the height in that point defined by f(x, y) so you end up suming the volume of each little parallelepiped, and so on, if  you take $\iint_{R}\rho(x,y) f(x,y)dA$ you multiply the volume of each little parallelepiped times the density at that point (mass = density*volume) then you end up suming the mass of each point bounded by your double integral, now you can think about the models in your calculus book.
