Question involving the double integral

From this wiki page multiple integral, the multiple integral is "an integral of a function over a two-dimensional region". But the caption of the image below states that "The double integral of the graphed function corresponds to the volume contained underneath the surface corresponding to the function." and volume is three-dimensional. What is this? Why is it contradicting? Am I missing something?

Note by Harry Lam
4 months, 2 weeks ago

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Well a single integral is the area under the curve so add a dimension to both

Jason Gomez - 4 months, 2 weeks ago

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Add a dimension to both what? The x and y axis? Now there are three dimensions - for finding the volume. But the definition says "the multiple integral is an integral of a function over a two-dimensional region" which contradicts this

Harry Lam - 4 months, 2 weeks ago

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Sorry I had meant add a dimension to the function, and add a dimension to the integral

The function was a curve previously, or 1 dimensional but now is 2d wavy sheet taking inputs from x and y and outputting as z

The single integral was the area under a 1 dimensional curve and now becomes the volume under a 2 dimensional wavy sheet

The volume under this is the double integral depending on what limits are taken

Jason Gomez - 4 months, 2 weeks ago

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@Jason Gomez Ok I see. But couldn't you do the same thing by making a solid of revolution to find the volume? Is that different?

Harry Lam - 4 months, 2 weeks ago

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@Harry Lam Solids of revolutions have one less degree of freedom than usual so they can be solved using single integrals itself ( but these single integrals are integrating areas into volumes, normally it integrates line segments into areas, look up on riemann sums if you didn’t understand the integrating line segments into area part )

Jason Gomez - 4 months, 2 weeks ago

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@Harry Lam I think that’s only possible when the figure has circular symmetry to exploit otherwise a change to the cylindrical co-ordinate system needs to be done ( I think that this is very hard to perform especially on random functions ) and also the limits will have to be well placed

I am very novice to double integrals, so I have very limited knowledge on them

Jason Gomez - 4 months, 2 weeks ago

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@Jason Gomez OK I see now. Thanks @Jasan Gomez for explaining it to me!

Harry Lam - 4 months, 2 weeks ago

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