How to change the order of integration in triple integrals?

I tried to make a graphic interpretation, but it seemed rather complex and didn't clarify the things. Moreover, I might not have much time if I had to solve this kind of problem on a test. So, I would be really grateful if someone could explain how to solve this problem efficiently.

$\begingroup$ Maybe it's worth bringing this thing to spherical coordinates instead? Or at least into something similar. $\endgroup$

$\begingroup$ @AlexeyBurdin: Spherical coordinates are definitely not going to make things easier in this case. And I don't think that's the point of the exercise either. $\endgroup$

2 Answers 2

The iterated integral on the left equals the triple integral $$ I = \iiint_D f \, dxdydz , $$ where $$ D = \< (x,y,z) \in \mathbf^3 : 0 \le x \le 1, \, 0 \le y \le 1, \, 0 \le z \le x^2+y^2 \> . $$ To write this as an iterated integral with $z$ outermost, we first figure out that the largest possible $z$ coordinate for a point in $D$ is $2$ , namely for the point $(x,y,z)=(1,1,2)$ . So the range of $z$ will be $0 \le z \le 2$ : $$ I = \int_^ \left( \iint_ f \, dxdy \right) dz , $$ where $D_z$ is the region in $\mathbf^2$ obtained by slicing through $D$ at a fixed height $z \in [0,2]$ : $$ D_z = \< (x,y) \in \mathbf^2 : 0 \le x \le 1, \, 0 \le y \le 1, \, x^2+y^2 \ge z \> . $$ Geometrically this is what you get when you take the unit square and remove a (quarter) disk with radius $\sqrt$ . Now writing a double integral over this region $D_z$ is slightly tricky, since it looks different depending on whether $z$ is less than or greater than $1$ (see case 1 and case 2 on Wolfram Alpha). So you'll need to split it, $$ I= \int_^ \left( \iint_ f \, dxdy \right) dz + \int_^ \left( \iint_ f \, dxdy \right) dz , $$ and then write $$ I= \int_^ \left( \int_^ \left( \int_^ f \, dx \right) \, dy \right) dz + \int_^ \left( \int_>^ \left( \int_>^ f \, dx \right) \, dy \right) dz , $$ where the curve $x=g(y)$ describes the left boundary of $D_y$ in case 1 (when $0\le z \le 1$ ): $$ g(y) = \begin \square, 0 \le y \le \square ,\\ \square, \square \le y \le 1 . \end $$ (I'm leaving it for you to figure out how to fill in the blanks in this final step, so that I don't spoil the whole exercise for you!)

$\begingroup$ Something wrong with the second term in $I$. When you integrate the inner $x$, it should not appear in the limit for $y$. The lower limit for $y$ should be $\sqrt<1-z>$. $\endgroup$

$\begingroup$ @Andrei: Oops! Of course there can be no $x$ there! (The dangers of solving in your head while typing instead of writing it down on paper first. ) It's actually $\sqrt$, but thanks for pointing it out anyway. Now it should hopefully be ok. $\endgroup$

$\begingroup$ @Bonrey: If you look at the Wolfram Alpha picture of $D_z$ in case 2, you see that the point in $D_z$ with the lowest value of the $y$ coordinate is the one in the lower right corner, $(x,y)=(1,\sqrt)$; it's the point where the circle $x^2+y^2=z$ (with $z \in [1,2]$) intersects the right edge of the unit square. $\endgroup$

$\begingroup$ Should the last system's blanks be filled that way (from the left to right, from the top to the bottom): $\sqrt,\ \sqrt,\ 0,\ \sqrt$ ? $\endgroup$