# Vector Calculus integration region.

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 runs outside the ellipse. Therefore I am not sure what is the region of the integration.

2)Furthermore, I believe Fubini's theorem requires $$g_1(x,y) for all $$x,y$$ in the region, but outside the ellipse the two upper and lower paraboloids swap i.e. $$g_2(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.

• 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. May 23, 2019 at 11:22
• 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 $x$. May 23, 2019 at 11:26
• yes it is 3y, sorry May 23, 2019 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.
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 $$0$$.