To evaluate the triple integral [tex]\(\iiint_E \sqrt{x^2 + z^2} \, dV\)[/tex], where [tex]\(E\)[/tex] is the region bounded by the paraboloid [tex]\(y = x^2 + z^2\)[/tex] and the plane [tex]\(y = 4\)[/tex], we can use cylindrical coordinates to simplify the calculation. Let's break this down step by step.
### Step 1: Convert to Cylindrical Coordinates
In cylindrical coordinates, the relationships between Cartesian coordinates [tex]\((x, y, z)\)[/tex] and cylindrical coordinates [tex]\((r, \theta, y)\)[/tex] are given by:
[tex]\[
x = r \cos \theta
\][/tex]
[tex]\[
z = r \sin \theta
\][/tex]
[tex]\[
r = \sqrt{x^2 + z^2}
\][/tex]
The volume element [tex]\(dV\)[/tex] in cylindrical coordinates is [tex]\(r \, dr \, d\theta \, dy\)[/tex].
### Step 2: Set Up the Integral
The integrand [tex]\(\sqrt{x^2 + z^2}\)[/tex] in cylindrical coordinates becomes [tex]\(\sqrt{r^2} = r\)[/tex]. So the integral to evaluate is:
[tex]\[
\iiint_E r \, r \, dr \, d\theta \, dy = \iiint_E r^2 \, dr \, d\theta \, dy
\][/tex]
### Step 3: Determine the Limits of Integration
Let's identify the bounds for [tex]\(r\)[/tex], [tex]\(\theta\)[/tex], and [tex]\(y\)[/tex]:
- [tex]\(r\)[/tex] ranges from [tex]\(0\)[/tex] to the radius at which the plane intersects the paraboloid. When [tex]\(y = 4\)[/tex], we have [tex]\(4 = r^2\)[/tex], which gives [tex]\(r = 2\)[/tex]. So, [tex]\(r\)[/tex] ranges from [tex]\(0\)[/tex] to [tex]\(2\)[/tex].
- [tex]\(\theta\)[/tex] ranges from [tex]\(0\)[/tex] to [tex]\(2\pi\)[/tex].
- [tex]\(y\)[/tex] ranges from the surface of the paraboloid [tex]\(y = r^2\)[/tex] to the plane [tex]\(y = 4\)[/tex].
Thus, our limits of integration are:
[tex]\[
r: 0 \text{ to } 2
\][/tex]
[tex]\[
\theta: 0 \text{ to } 2\pi
\][/tex]
[tex]\[
y: r^2 \text{ to } 4
\][/tex]
### Step 4: Write the Triple Integral
Substituting these limits and the integrand into the triple integral, we get:
[tex]\[
\iiint_E r^2 \, dy \, dr \, d\theta = \int_0^{2\pi} \int_0^2 \int_{r^2}^4 r^2 \, dy \, dr \, d\theta
\][/tex]
### Step 5: Evaluate the Integral
First, we integrate with respect to [tex]\(y\)[/tex]:
[tex]\[
\int_{r^2}^4 r^2 \, dy = r^2 [y]_{r^2}^4 = r^2 (4 - r^2)
\][/tex]
Now our integral becomes:
[tex]\[
\int_0^{2\pi} \int_0^2 r^2 (4 - r^2) \, dr \, d\theta
\][/tex]
Evaluate the inner integral with respect to [tex]\(r\)[/tex]:
[tex]\[
\int_0^2 r^2 (4 - r^2) \, dr = \int_0^2 (4r^2 - r^4) \, dr
\][/tex]
This can be separated into two integrals:
[tex]\[
\int_0^2 4r^2 \, dr - \int_0^2 r^4 \, dr
\][/tex]
Evaluate each integral:
[tex]\[
\int_0^2 4r^2 \, dr = 4 \int_0^2 r^2 \, dr = 4 \left[ \frac{r^3}{3} \right]_0^2 = 4 \cdot \frac{8}{3} = \frac{32}{3}
\][/tex]
[tex]\[
\int_0^2 r^4 \, dr = \left[ \frac{r^5}{5} \right]_0^2 = \frac{32}{5}
\][/tex]
Combine the results:
[tex]\[
\frac{32}{3} - \frac{32}{5} = \frac{160}{15} - \frac{96}{15} = \frac{64}{15}
\][/tex]
Now the integral becomes:
[tex]\[
\int_0^{2\pi} \frac{64}{15} \, d\theta
\][/tex]
Finally, integrate with respect to [tex]\(\theta\)[/tex]:
[tex]\[
\int_0^{2\pi} \frac{64}{15} \, d\theta = \frac{64}{15} \left[ \theta \right]_0^{2\pi} = \frac{64}{15} \cdot 2\pi = \frac{128\pi}{15}
\][/tex]
Therefore, the value of the integral is:
[tex]\[
\boxed{\frac{128\pi}{15}}
\][/tex]
