How to know whether to choose the x-bound or the y-bound for this triple integral

Question

In my textbook for calculus 3, I have been working on example of the triple integral. Though I do know polar, cylindrical, spherical coordinates, this section of the book expects you to work with Cartesian coordinates. Anyways the problem is given as follows:

Evaluate a function $f(x, y, z) = x$, in the first octant above $z = y^{2}$ and below $z = 8 - 2x^{2} - y^{2}$.

I began by defining my $z$ bounds:

$y^{2} \leq z \leq 8 - 2x^{2} - y^{2}$.

Next, I decided that I would substitute the the lower bound of $z$ into the upper bound to find the intersection. My understanding is that once I have found this intersection, I can use it as the projection onto the $xy$ plane. I find

$x^{2} + y^{2} = 4$

I then define my $y$ bounds:

$0 \leq y \leq \sqrt{4 - x^{2}}$

This leaves my $x$ bounds as:

$0 \leq x \leq 2$

For some reason, evaluating the integral with these bounds turns out to be much more difficult than if I had sort of swapped my bounds for $x$ and $y$. My gut tells me that this is a result of the $x$ within the function being integrated, yet I cannot convince myself why (and that might be entirely wrong). I was wondering if anybody could explain the reason why the bounds I chose are less practical than the alternative. Also, I am seriously struggling with bounding equations and I was wondering if anybody had anything good I might want to read or perhaps a set of problems that builds up understanding very well. For reference, I am working out Rogawski, and my experience with this book is that just as I start to understand it a bit more, I run out of problems...

Answer 1

I was able to evaluate (I think) the integral given your bounds, I suspect you just missed a simplification step somewhere, possibly involving the square root. \begin{align*} \iiint f(x,y,z) \ \mathrm{d} z \ \mathrm{d} y \ \mathrm{d} x &= \int_0^2 \int_0^{\sqrt{4-x^2}} \int^{8-2x^2-y^2}_{y^2} x \ \ \mathrm{d} z \ \mathrm{d} y \ \mathrm{d} x \\ & = \int_0^2 \int_0^{\sqrt{4-x^2}} \left[ xz \right]^{z=8-2x^2-y^2}_{z=y^2} \ \ \mathrm{d} y \ \mathrm{d} x \\ & = \int_0^2 \int_0^{\sqrt{4-x^2}} 8x-2x^3-xy^2 - xy^2 \ \ \mathrm{d} y \ \mathrm{d} x \\ & = \int_0^2 \int_0^{\sqrt{4-x^2}} 8x-2x^3 - 2xy^2 \ \ \mathrm{d} y \ \mathrm{d} x \\ & = \int_0^2 \left[ (8x-2x^3)y - \frac{2}{3}xy^3 \right]^{y=\sqrt{4-x^2}}_{y=0} \ \ \mathrm{d} x \\ & = \int_0^2 2x(4-x^2)\sqrt{4-x^2} - \frac{2}{3}x(4-x^2)^{3/2} \ \ \mathrm{d} x \\ & = \int_0^2 2x(4-x^2)^{3/2} - \frac{2}{3}x(4-x^2)^{3/2} \ \ \mathrm{d} x \\ & = \frac{4}{3}\int_0^2 x(4-x^2)^{3/2} \ \ \mathrm{d} x \\ & = \frac{\color{blue}{2}}{3} \int_0^2 \color{blue}{2}x(4-x^2)^{3/2} \ \ \mathrm{d} x \\ & = \frac{2}{3}\cdot\frac{2}{5}\left[ - (4-x^2)^{5/2} \right]^2_0 \qquad \qquad \color{blue}{(\mbox{Note: done via substitution)}}\\ & = \frac{4}{15}\left[ 0 + 4^{5/2} \right] \\ & = \frac{4}{15} \cdot 2^5 \\ & = \frac{128}{15} \end{align*}

But you are correct, we could attempt this integral swapping the $x$ and $y$ bounds, from $x^2+y^2=4$ we have $0\le x \le \sqrt{4-y^2}$, and $0 \le 2 \le y$. I'll do the integral again using this method.

\begin{align*} \iiint f(x,y,z) \ \mathrm{d} z \ \mathrm{d} x \ \mathrm{d} y &= \int_0^2 \int_0^{\sqrt{4-y^2}} \int^{8-2x^2-y^2}_{y^2} x \ \ \mathrm{d} z \ \mathrm{d} x \ \mathrm{d} y \\ & = \int_0^2 \int_0^{\sqrt{4-y^2}} \left[ xz \right]^{z=8-2x^2-y^2}_{z=y^2} \ \ \mathrm{d} x \ \mathrm{d} y \\ & = \int_0^2 \int_0^{\sqrt{4-y^2}} 8x-2x^3-xy^2 - xy^2 \ \ \mathrm{d} x \ \mathrm{d} y \\ & = \int_0^2 \int_0^{\sqrt{4-y^2}} 8x-2x^3 - 2xy^2 \ \ \mathrm{d} x \ \mathrm{d} y \\ & = \int_0^2 \left[ 4x^2-\frac{1}{2}x^4 - x^2y^2\right]^{x=\sqrt{4-y^2}}_{x=0} \ \ \mathrm{d} y \\ & = \int_0^2 4(4-y^2) -\frac{1}{2}(4-y^2)^2 -(4-y^2)y^2 \ \ \mathrm{d} y \\ & = \frac{1}{2} \int_0^2 16 - 8 y^2 + y^4 \ \ \mathrm{d} y \qquad \qquad \color{grey}{\mbox{(done in W|A)}} \\ & = \frac{1}{2}\left[16y - \frac{8}{3}y^3 + \frac{1}{5}y^5\right]^2_0 \\ & = \frac{1}{2}\left(16(2) - \frac{8}{3}(2)^3 + \frac{1}{5}\cdot 2^5 \right) \\ & = \frac{128}{5} \end{align*}

Now, this second method leaves you just with a polynomial to integrate rather than a substitution integral, but I'd say they are roughly equivalent in difficulty. Sometimes you simply can't integrate a given function using a particular order, and must swap. For example, integrating the function $f(x,y) = \exp(x^2)$ over the triangle in the first quadrant bounded by the $x$-axis, $y=x$ and $x=1$, is simply impossible using the order $$ \iint f(x,y) \ \ \mathrm{d} x \ \mathrm{d} y = \int_{y=0}^{y=1} \int_{x=y}^{x=1} \mathrm{e}^{x^2} \ \ \mathrm{d} x \ \mathrm{d} y, $$ as you probably know. However if we swap the order, $$ \iint f(x,y) \ \ \mathrm{d} y \ \mathrm{d} x = \int_{x=0}^{x=1} \int_{y=0}^{y=x} \mathrm{e}^{x^2} \ \ \mathrm{d} y \ \mathrm{d} x, $$ we obtain $$ \iint f(x,y) \ \ \mathrm{d} y \ \mathrm{d} x = \int_{0}^{1} x\mathrm{e}^{x^2} \ \mathrm{d} x, $$ which is a simple substitution integral.

Final advice would be to keep practising, remember your substitution technique from your earlier studies, and maybe have a look at these notes and practise some double integrals where you swap the order - I always find this guy's site helpful for practise problems.