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anonymous
 one year ago
What exactly is the difference between a double integral and a triple integral in terms of finding volume? Like, are there solids that require the use of a triple integral instead of a double integral to evaluate, and can someone give me an example of one?
anonymous
 one year ago
What exactly is the difference between a double integral and a triple integral in terms of finding volume? Like, are there solids that require the use of a triple integral instead of a double integral to evaluate, and can someone give me an example of one?

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2In cartesian coordinates, Volume between two surfaces \(f(x,y)\) and \(g(x,y)\) is given by : \[V ~=~ \iint\color{blue}{\int\limits_{f(x,y)}^{g(x,y)}~1~dz}dydx ~=~ \iint \color{blue}{g(x,y)f(x,y)}\,dydx\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Notice that the innermost integral in triple integral evaluates to the integrand in double integral. So, to my knowledge, the double integral is just a formula which can always be produced from the triple integral. They are same.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, that's what I thought. But what about cases where the two functions cannot be put into a form of \(z=f(x, y)\), where z isn't or cannot be defined explicitly in terms of x any y? Is that when nonrectangular coordinate systems come in or would triple integrals help with that at all?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I don't see how a triple integral "helps" in reducing the complexity. Because you can always change back and forth between them by evaluating the trivial inner most integral : \[\large \color{blue}{\int\limits_{f(x,y)}^{g(x,y)}\,1\,dz} ~~ =~~ \color{blue}{g(x,y)f(x,y)} \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2triple integral simply reduces to double integral after you evaluate that inner most integral, so...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, I think I get it. I was under the impression that the integrand of the triple integral when calculating volume could be something other than 1, and I think that's where I got confused.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for clarifying things for me!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Yes, the integrand must be 1 when you use triple integral for computing volume. Its similar to how you use 1 as integrand for representing area between two curves using double integral : \[A = \int\color{blue}{\int\limits_{f(x)}^{g(x)}\,1\,dy}dx~~=~~\int\color{blue}{g(x)f(x)}\,dx\] Notice that the blue parts are same.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, thanks again. I've been selfstudying multivariable calculus, so it's a bit odd when I run into a problem like this and not have a teacher to clarify.
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