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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?

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  1. ganeshie8
    • one year ago
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    In cartesian coordinates, Volume between two surfaces \(f(x,y)\) and \(g(x,y)\) is given by : \[V ~=~ \iint\color{blue}{\int\limits_{f(x,y)}^{g(x,y)}~1~dz}dydx ~=~ \iint \color{blue}{g(x,y)-f(x,y)}\,dydx\]

  2. ganeshie8
    • one year ago
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    Notice 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.

  3. anonymous
    • one year ago
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    Right, 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 non-rectangular coordinate systems come in or would triple integrals help with that at all?

  4. ganeshie8
    • one year ago
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    I 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 : \[\large \color{blue}{\int\limits_{f(x,y)}^{g(x,y)}\,1\,dz} ~~ =~~ \color{blue}{g(x,y)-f(x,y)} \]

  5. ganeshie8
    • one year ago
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    triple integral simply reduces to double integral after you evaluate that inner most integral, so...

  6. anonymous
    • one year ago
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    Alright, 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.

  7. anonymous
    • one year ago
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    Thanks for clarifying things for me!

  8. ganeshie8
    • one year ago
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    Yes, 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.

  9. anonymous
    • one year ago
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    Alright, thanks again. I've been self-studying multivariable calculus, so it's a bit odd when I run into a problem like this and not have a teacher to clarify.

  10. ganeshie8
    • one year ago
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    np

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