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Volume of a Solid- 3D Calculus
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[QUOTE="haruspex, post: 5460951, member: 334404"] No, a volume integral would just be ∫∫∫dxdydz, no need for a function. But in this case fridakahlo is effectively assuming the z integral has been done, leaving either ∫∫fdxdy or ∫∫dydx. [USER=574106]@fridakahlo[/USER] , yes you can do the x integral before y, but you would need to split the outermost integral into two ranges, y<2.5 and y>2.5. [/QUOTE]
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Volume of a Solid- 3D Calculus
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