Evaluate the integral by changing to cylindrical coordinates.

Integral2-2 integralsquareroot 4 – y2-squareroot 4 – y2 integral2squareroot x2 + y2 xz dz dx dy

integral3-3 integralsquareroot 9 – x20 integral9-x2-y20 squareroot x2 + y2 dz dy dx

The Correct Answer and Explanation is:

We are given a triple integral, and the task is to evaluate it by converting to cylindrical coordinates. Let’s first decode and understand the expression clearly.

Interpreting the Integral:

The given expression seems to be:

$$
\int_{-2}^{2} \int_{-\sqrt{4 – y^2}}^{\sqrt{4 – y^2}} \int_{0}^{2\sqrt{x^2 + y^2}} xz \, dz \, dx \, dy
$$

However, another version follows:

$$
\int_{-3}^{3} \int_{-\sqrt{9 – x^2}}^{\sqrt{9 – x^2}} \int_{0}^{\sqrt{9 – x^2 – y^2}} \sqrt{x^2 + y^2} \, dz \, dy \, dx
$$

We’ll evaluate the first integral by converting to cylindrical coordinates.

Step 1: Change to Cylindrical Coordinates

Cylindrical coordinates are:

$x = r \cos\theta$
$y = r \sin\theta$
$z = z$
$dx\,dy = r\,dr\,d\theta$

In the integrand $xz$, we substitute:

$$
xz = (r \cos\theta)z
$$

Step 2: Analyze the Region

The limits:

Outer: $x \in [-2, 2]$
Middle: $y \in [-\sqrt{4 – x^2}, \sqrt{4 – x^2}]$
→ Together define a disk of radius 2 in the xy-plane.
Inner: $z \in [0, 2\sqrt{x^2 + y^2}]$
→ This is $z \in [0, 2r]$ in cylindrical coordinates.

Step 3: Set up the Cylindrical Integral

Now rewrite the integral:

$$
\int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{2r} (r \cos\theta)z \cdot r \, dz \, dr \, d\theta
$$

Simplify the integrand:

$$
r^2 \cos\theta \cdot z
$$

So the integral becomes:

$$
\int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{2r} r^2 \cos\theta \cdot z \, dz \, dr \, d\theta
$$

Step 4: Evaluate the Integral

Innermost integral:

$$
\int_{0}^{2r} z \, dz = \left[\frac{1}{2}z^2\right]_0^{2r} = \frac{1}{2}(4r^2) = 2r^2
$$

Middle integral:

$$
\int_{0}^{2} r^2 \cdot 2r^2 \, dr = 2 \int_{0}^{2} r^4 \, dr = 2 \cdot \left[\frac{r^5}{5}\right]_0^2 = 2 \cdot \frac{32}{5} = \frac{64}{5}
$$

Outer integral:

$$
\int_{0}^{2\pi} \cos\theta \, d\theta = 0
$$

Final Answer:

Since the outermost integral equals 0, the value of the entire triple integral is:

$$
\boxed{0}
$$

300-Word Explanation:

This problem involves evaluating a triple integral using cylindrical coordinates. The original region of integration defines a circular disk of radius 2 in the xy-plane, and a height that depends on the distance from the origin, namely $z = 2\sqrt{x^2 + y^2}$. Because the integrand contains $xz$, which becomes $r \cos\theta \cdot z$ in cylindrical coordinates, we observe the presence of a trigonometric term, $\cos\theta$.

Cylindrical coordinates are useful in this case because the symmetry of the region (a disk) matches the natural structure of cylindrical systems. By expressing $x = r\cos\theta$, $y = r\sin\theta$, and $dx\,dy = r\,dr\,d\theta$, we transform the region into polar bounds, greatly simplifying the integration process.

After converting the triple integral, the integrand becomes $r^2 \cos\theta \cdot z$, and the limits become $r \in [0, 2]$, $\theta \in [0, 2\pi]$, and $z \in [0, 2r]$. Integrating with respect to $z$ gives $2r^2$. Continuing with respect to $r$, the integral becomes $\frac{64}{5}$. However, when integrating the outermost function $\cos\theta$ from $0$ to $2\pi$, we get zero because $\cos\theta$ is symmetric about the y-axis and cancels out over a full period.

This cancellation is key—despite nonzero inner integrals, the final result is zero due to symmetry. Therefore, the value of the integral is: