Algorithmic approach to double/triple integrals

[oreosama]

I feel overwhelmed with something that should be capable of being explained very simply I think.

Let's say you're getting thrown random questions involving surfaces/shapes creating boundries in ℝ3. Whats your step by step process in finding whether you want to do a double integral versus triple, finding your limits and the actual function youre going to integrate over, etc.

It seems silly but I guess I just blindly manipulate the information in hopes that the answer jumps out to me-- I can solve through the problems but I certainly can't say I understand what I'm doing other than just going through the motions.

 

[Blibbler]

I just found this on another forum:

Long story short: You can use a triple integral with a unit integrand (∫∫∫dV) to find the volume of any closed figure in R3. The boundary of the region is encoded in the limits of integration. You can use double integrals of the form ∫∫f(x1,x2)dS to find the volume of a region in R3 provided that the region is partially bounded by the surface x3=f(x1,x2). And lastly you can use single integrals to find volumes of regions of R3 bounded by surfaces of revolution.

Loosely speaking the fewer integrations in your volume computation, the more picky you have to be about the region.

Seems to answer the question.

 

[oreosama]

the triple unit integrand stuff shows a pretty direct connection between triple and double integrals which is cool I guess

not all volumes I deal with involve just unit integrands though. if you have some volume, and you triple integrate through there could be chunks missing if I understand what I'm talking about. this some volume should be what youre entirely integrating over, right?


Use a triple integral and spherical coordinates to find the volume of the solid that is
bounded by the graphs

z = √(x^2+y^2) and x^2 + y^2 + z^2 = 9


since if you draw this out everything is encompassed by the sphere that we want to integrate you'd do a triple integral over z= √(9 - x^2 - y^2) (converted to spherical), right?

 

[SammyS]

the triple unit integrand stuff shows a pretty direct connection between triple and double integrals which is cool I guess

not all volumes I deal with involve just unit integrands though. if you have some volume, and you triple integrate through there could be chunks missing if I understand what I'm talking about. this some volume should be what you're entirely integrating over, right?


Use a triple integral and spherical coordinates to find the volume of the solid that is
bounded by the graphs

z = √(x^2+y^2) and x^2 + y^2 + z^2 = 9

since if you draw this out everything is encompassed by the sphere that we want to integrate you'd do a triple integral over z= √(9 - x^2 - y^2) (converted to spherical), right?

The most important algorithm, is to use your head. Experience helps. Sketching the region helps. Much depends on the integrand. Are you merely finding the volume, or are you integrating some function over a three dimensional region.

It may be that the integration itself -- that is to say, the way the limits of integration interact with the anti-derivative of the integrand -- works out much nicer with one scheme as compared to another.

For the example you give, each the following methods works pretty well:

1. Using spherical coordinates:

2. Using cylindrical coordinates:

3. Volume of revolution:​

A couple of other approaches also come to mind.