Evaluate $ \int_0^\sqrt2\int_0^{3y}\int_{x^2+3y^2}^{8-x^2-y^2}dzdxdy$. The method given in the answer booklet was to calculate the integrals one at a time and get a numerical answer and it is quite straightforward.

1)However I am not sure what the region of integration is. The two paraboloids intersect along a curve which when projected on the $xy-$plane is an ellipse of equation $x^2+2y^2=4$, obtained by setting $z_1=8-x^2-y^2,z_2=x^2+3y^2$ equal to each other. However the region $0<x<3y,0<y<\sqrt2$ runs outside the ellipse. Therefore I am not sure what is the region of the integration.

2)Furthermore, I believe Fubini's theorem requires $g_1(x,y)<z<g_2(x,y)$ for all $x,y$ in the region, but outside the ellipse the two upper and lower paraboloids swap i.e. $g_2(x,y)<z<g_1(x,y)$. So should we compute another integral separately for the region outside the ellipse? In which case the answer in the booklet is wrong.

3)Lastly, is it true that for two functions $z_1=f(x,y)$ and $z_2=g(x,y)$, to find the projection of the surface on the xy-plane we always set them equal to each other. Why is that? And if instead $z_1^2=f(x,y)$ do we set $z_2=\pm\sqrt{z_1}$ and obtain two projections on the xy-plane?Why again?

I greatly appreciate all your responses answering my three questions as it will greatly enlighten me on some basic ideas of vector calculus. Thank you very much.

  • $\begingroup$ Are you certain that you have the right bounds and the right order of $dz\,dx\,dy$? Because letting the $dx$ integral have $x$ in the boundary is just strange. $\endgroup$ – Arthur May 23 at 11:22
  • $\begingroup$ Or perhaps the inner upper boundary is not $3x$ (but a function of $y$) since the outer boundaries seem to make more sense for $y$ than $x$. $\endgroup$ – StackTD May 23 at 11:26
  • $\begingroup$ yes it is 3y, sorry $\endgroup$ – johnson May 23 at 12:12

You are correct in the sense that with these given boundaries, there does not seem to correspond a 'logical' region as you would expect, such as "the region bounded by the paraboloids and [some extra restrictions on the region in the $xy$-plane]". This makes it hard to convert the given boundaries back to a region which is easy to describe/explain visually - so without simply referring to the given boundaries.

However: the boundaries are what they are. If the intention was to describe a 'logical region' in the sense above, then perhaps there's a mistake in the boundaries. But perhaps that was never the idea and this is just an exercise to practice on evaluating an iterated integral.

Note that you would use Fubini to convert a triple or double integral over some region to an iterated integral - but you don't have to do that here because the iterated integrals are already given!

Compare it to evaluating: $$\int_0^\color{red}{3} \int_{x^2}^{2x} \,\mbox{d}y\,\mbox{d}x \tag{$\star$}$$ If you try to reason back to the region of integration, you may think they want you to find an area between a parabola and a line. However, you would expect the upper outer boundary to be $2$ instead of $3$! Or, if you do continue after $x=2$ and you still want to find the area between the curves, you would split because after $x=2$ the line is below the parabola (instead of the other way around). But perhaps $(\star)$ was the intended iterated integral and it evaluates to $0$.


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