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To evaluate the given triple integral

\[
\iiint_E 8x \, dV,
\]
where \( E \) is the region bounded by the paraboloid \( x = 7y^2 + 7z^2 \) and the plane \( x = 7 \), we can follow these steps:

### Step 1: Set up the region \( E \)
The region \( E \) is enclosed between the surface \( x = 7(y^2 + z^2) \) (a paraboloid) and the plane \( x = 7 \). The equation of the paraboloid can be rewritten as:

\[
x = 7(y^2 + z^2),
\]
and the plane \( x = 7 \) represents the upper boundary. This is a natural fit for cylindrical coordinates.

### Step 2: Use cylindrical coordinates
In cylindrical coordinates, the variables are transformed as follows:
- \( y = r\cos\theta \),
- \( z = r\sin\theta \),
- \( y^2 + z^2 = r^2 \),
and the volume element \( dV \) becomes \( r \, dx \, dr \, d\theta \).

Thus, the equation for the paraboloid becomes:

\[
x = 7r^2,
\]
and the upper bound is \( x = 7 \).

### Step 3: Set up the limits of integration
In cylindrical coordinates:
- \( r \) ranges from 0 to 1 (since the paraboloid intersects the plane at \( r = 1 \), because \( 7r^2 = 7 \) when \( r = 1 \)),
- \( \theta \) ranges from 0 to \( 2\pi \) (since it's a full rotation around the z-axis),
- \( x \) ranges from \( 7r^2 \) (the paraboloid) to 7 (the plane).

### Step 4: Express the integral in cylindrical coordinates
The triple integral becomes:

\[
\int_0^{2\pi} \int_0^1 \int_{7r^2}^{7} 8x \, r \, dx \, dr \, d\theta.
\]

### Step 5: Compute the integral
First, integrate with respect to \( x \):

\[
\int_{7r^2}^7 8x \, dx = 4x^2 \Big|_{7r^2}^7 = 4(7^2) - 4(7r^2)^2 = 196 - 196r^4.
\]

Now, substitute this into the remaining double integral:

\[
\int_0^{2\pi} \int_0^1 r(196 - 196r^4) \, dr \, d\theta = 196 \int_0^{2\pi} \int_0^1 (r - r^5) \, dr \, d\theta.
\]

First, compute the \( r \)-integral:

\[
\int_0^1 (r - r^5) \, dr = \left[ \frac{r^2}{2} - \frac{r^6}{6} \right]_0^1 = \frac{1}{2} - \frac{1}{6} = \frac{1}{3}.
\]

Now, compute the \( \theta \)-integral:

\[
196 \cdot \frac{1}{3} \cdot \int_0^{2\pi} d\theta = 196 \cdot \frac{1}{3} \cdot 2\pi = \frac{392\pi}{3}.
\]

### Final Answer:
The value of the triple integral is:

\[
\boxed{\frac{392\pi}{3}}.
\]

---

Would you like further details or clarifications on any step? Here are some related questions you might find helpful:

1. How do cylindrical coordinates help simplify triple integrals?
2. What are the steps for converting a given integral to cylindrical coordinates?
3. Why is the region bounded by the paraboloid and plane suited for cylindrical coordinates?
4. How would you approach this problem if the bounds were different or in Cartesian coordinates?
5. Could the same approach be used for other types of surfaces, such as cones or spheres?

**Tip:** When handling triple integrals over symmetric regions, switching to cylindrical or spherical coordinates can often simplify the process significantly!

Evaluate the triple integral ∭_E 8x dV, where E is bounded by the paraboloid x = 7y² + 7z² and the plane x = 7.

Learn how to evaluate the triple integral ∭_E 8x dV, where E is bounded by the paraboloid x = 7y² + 7z² and the plane x = 7, using cylindrical coordinates. This detailed solution covers key concepts in multivariable calculus, including coordinate transformations and integration limits. Suitable for university-level students.

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