Evaluate the following integral in cylindrical coordinates.

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Part 1

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​(Simplify your answer. Type an exact​ answer, using 

piπ 

as​ needed.)

We are asked to evaluate the following triple integral in cylindrical coordinates:

\[
\int_{0}^{3} \int_{0}^{\sqrt{2}} \int_{x}^{\sqrt{16 - x^2}} e^{-x^2 - y^2} \, dy \, dx \, dz
\]

### Step 1: Express the integral in cylindrical coordinates

In cylindrical coordinates, we have the following relationships:

- \(x = r \cos \theta\)
- \(y = r \sin \theta\)
- \(z = z\) (unchanged)
- The area element \(dx\,dy\) becomes \(r \, dr \, d\theta\) due to the Jacobian in cylindrical coordinates.

Moreover, \( x^2 + y^2 = r^2 \), and thus the exponential term \(e^{-x^2 - y^2}\) transforms to \(e^{-r^2}\).

### Step 2: Boundaries for cylindrical coordinates

We need to convert the limits of integration:

- \(z\) runs from 0 to 3.
- The \(xy\)-region is circular, as \(x^2 + y^2 \leq 16\), so \(r\) ranges from 0 to 4 (since \(\sqrt{16} = 4\)).
- The angle \(\theta\) will range from 0 to \(2\pi\) for a full circular region.

Thus, the integral becomes:

\[
\int_0^3 \int_0^{4} \int_0^{2\pi} r e^{-r^2} \, d\theta \, dr \, dz
\]

### Step 3: Evaluate the integral over \(\theta\)

The \(\theta\)-integral is straightforward:

\[
\int_0^{2\pi} d\theta = 2\pi
\]

This simplifies the integral to:

\[
2\pi \int_0^3 \int_0^4 r e^{-r^2} \, dr \, dz
\]

### Step 4: Evaluate the radial integral

We now focus on the radial part:

\[
\int_0^4 r e^{-r^2} \, dr
\]

This integral can be solved using substitution. Let \( u = r^2 \), so \( du = 2r \, dr \), or \( \frac{du}{2} = r \, dr \). The limits for \( u \) will go from 0 to \( 16 \) as \( r \) ranges from 0 to 4.

Thus, the radial integral becomes:

\[
\frac{1}{2} \int_0^{16} e^{-u} \, du
\]

The integral of \(e^{-u}\) is simply \(-e^{-u}\), so:

\[
\frac{1}{2} \left[ -e^{-u} \right]_0^{16} = \frac{1}{2} \left( -e^{-16} + e^0 \right) = \frac{1}{2} \left( 1 - e^{-16} \right)
\]

### Step 5: Putting it all together

Now, the remaining integral is:

\[
2\pi \int_0^3 \frac{1}{2} \left( 1 - e^{-16} \right) \, dz
\]

This simplifies to:

\[
\pi \left( 1 - e^{-16} \right) \int_0^3 dz = \pi \left( 1 - e^{-16} \right) \cdot 3
\]

Thus, the final answer is:

\[
\boxed{3\pi \left( 1 - e^{-16} \right)}
\]

Would you like more details on any part of this solution? Here are some related questions to practice:

1. How do you compute integrals involving cylindrical coordinates?
2. What are common transformations for triple integrals into cylindrical coordinates?
3. Why does the Jacobian in cylindrical coordinates introduce an extra \(r\)?
4. How do you handle substitution in exponential integrals like \(\int r e^{-r^2} \, dr\)?
5. What is the significance of boundary conditions when converting to cylindrical coordinates?

**Tip**: Always check the region of integration carefully when switching to cylindrical or spherical coordinates—it can greatly simplify the integral!

Evaluate the following integral in cylindrical coordinates: 3∫ 022∫0 16−x2∫xe−x2−y2dy dx dz (Simplify your answer. Type an exact answer, using pi as needed.)

Learn how to solve a triple integral by converting it into cylindrical coordinates. This problem involves exponential functions and coordinate transformations. Suitable for university-level students studying multivariable calculus.

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