Change the order of integration: $\int_0^1 \int_0^1 \int_{x^2}^1 12 xz \exp(z y^2) \mathrm dy \mathrm dx \mathrm dz$

Question

Consider the following integral:

\begin{equation} \int_0^1 \int_0^1 \int_{x^2}^1 12 xz \exp(z y^2) \ \mathrm dy \ \mathrm dx \ \mathrm dz. \end{equation}

It has been said that the integral cannot be calculated, and one should first change the order of integration.

Any help concerning this problem is much appreciated. I cannot get to a final calculation, I am assuming I am not using the correct order of integration.

I try to change the integration from dy to dx and switch the bounds from 0 to 1 and sqrt(y) to 1 ... after doing this I integrate normally but either way I am still unable to solve the question.

Answer 1

In your last paragraph, it sounds like you're changing the integral to $$ \int _ 0 ^ 1 \int _ 0 ^ 1 \int _ { \sqrt y } ^ 1 \, \cdots \, \mathrm d x \, \mathrm d y \, \mathrm d z \text . $$ But this is incorrect. You're right that $ x = \sqrt { \mathstrut y } $ is one of the bounds (coming from $ y = x ^ 2 $ in the original integral, and using that $ x \geq 0 $ in the original integral to choose $ x = \sqrt { \mathstrut y } $ and reject $ x = - \sqrt { \mathstrut y } $). But $ x = 1 $ is not a correct bound. (Maybe this is enough for you to fix it; if not, then read on.)

Ignoring $ z $ for the moment, the original integral has four bounds: $ y = x ^ 2 $, $ y = 1 $, $ x = 0 $, and $ x = 1 $. But one of these is what I call a fake bound: it appears in an outer integral, but it comes from equating the two bounds in an inner integral, rather than from an actual boundary curve (or surface) of the region of integration. In this case, equating the bounds for $ y $ in the inner integral gives $ x ^ 2 = 1 $, so $ x = 1 $ (one of the solutions to this) is a fake bound. You really only have $ y = x ^ 2 $, $ y = 1 $, and $ x = 0 $.

Solving $ y = x ^ 2 $ for $ x $ (and remembering that $ x \geq 0 $), you get $ x = \sqrt { \mathstrut y } $. So the bounds for $ x $ are $ x = \sqrt { \mathstrut y } $ and $ x = 0 $. Then one of the bounds for $ y $ is $ y = 1 $, and you get the other bound for $ y $ by equating the bounds for $ x $ in the inner integral to get $ \sqrt { \mathstrut y } = 0 $, so $ y = 0 $. Restoring $ z $ (which wasn't involved in the rest of this), the correct integral is $$ \int _ 0 ^ 1 \int _ 0 ^ 1 \int _ 0 ^ { \sqrt y } \, \cdots \, \mathrm d x \, \mathrm d y \, \mathrm d z \text . $$ (But if you want to change this again, remember that $ y = 0 $ is a fake bound.)

Now when you do the inner integral with respect to $ x $, you'll get an integral with respect to $ y $ that you can actually do.

Answer 2

Notice that the region $$0\leq x^2\leq y\leq1$$ is the region $$0\leq x\leq\sqrt y\leq 1$$ on the $xy$-plane. Therefore, we can change the iteration of the integral $$\bbox[10px, yellow, border: 2px solid blue]{\begin{align}x^2\leq y\leq 1\\0\leq x\leq1\\0\leq z\leq 1\end{align}}\to \bbox[10px, yellow, border: 2px solid blue]{\begin{align}0\leq x\leq \sqrt y\\0\leq y\leq1\\0\leq z\leq 1\end{align}}$$ The evaluation of the integral with the new bounds is done by dfnu in the comments. It is correct.