Let's break down the given problem, which involves converting the triple integral from Cartesian to cylindrical coordinates and then evaluating the result.

Original Integral:

$\int_0^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{-(x^2 + y^2)}^{x^2 + y^2} 21xy^2 \, dz \, dy \, dx$

Step 1: Convert to cylindrical coordinates

In cylindrical coordinates:

• $x = r \cos \theta$
• $y = r \sin \theta$
• $z = z$
• The Jacobian determinant for converting from Cartesian to cylindrical coordinates is $r$, which means that the volume element $dx \, dy \, dz$ becomes $r \, dz \, dr \, d\theta$.

The integral $x^2 + y^2 = r^2$, and the bounds will adjust accordingly:

• $x^2 + y^2$ becomes $r^2$.
• The limits on $z$ remain the same, but we rewrite them in terms of $r$: $z \in [-r^2, r^2]$.

Step 2: Rewriting the integral in cylindrical coordinates

The integrand $21xy^2$ becomes: $21xy^2 = 21r^3 \cos \theta \sin^2 \theta$

Now, we can rewrite the integral in cylindrical coordinates: $\int_0^{2\pi} \int_0^1 \int_{-r^2}^{r^2} 21r^3 \cos \theta \sin^2 \theta \, dz \, r \, dr \, d\theta$

Step 3: Simplifying the integral

First, evaluate the $z$-integral. The limits for $z$ are from $-r^2$ to $r^2$, and the integrand is independent of $z$, so the result is just the length of the interval $2r^2$: $\int_{-r^2}^{r^2} dz = 2r^2$

Thus, the integral simplifies to: