Evaluate the triple integral.
 

E
7x dV,
 where E is bounded by the paraboloid 
x = 2y2 + 2z2
 and the plane 
x = 2

We are asked to evaluate the triple integral \( \iiint_E 7x \, dV \), where the region \( E \) is bounded by the paraboloid \( x = 2y^2 + 2z^2 \) and the plane \( x = 2 \).

### Step 1: Describe the region \( E \)

The region \( E \) is defined by:
1. The surface \( x = 2y^2 + 2z^2 \), a paraboloid opening along the positive \( x \)-axis.
2. The plane \( x = 2 \), which acts as an upper boundary for \( x \).

We convert the equation \( x = 2y^2 + 2z^2 \) into cylindrical coordinates, where:
- \( y = r\cos(\theta) \)
- \( z = r\sin(\theta) \)
- \( x = x \) (unchanged)

The equation of the paraboloid becomes:
\[
x = 2r^2
\]
The plane \( x = 2 \) remains unchanged.

### Step 2: Set up the limits of integration in cylindrical coordinates

1. **\( x \)-bounds**: \( x \) ranges from the paraboloid \( x = 2r^2 \) to the plane \( x = 2 \).
2. **\( r \)-bounds**: The maximum value of \( r \) occurs when \( x = 2 \), i.e., from \( 2r^2 = 2 \) which gives \( r = 1 \). So \( r \) ranges from 0 to 1.
3. **\( \theta \)-bounds**: Since the region is symmetric about the \( z \)-axis, \( \theta \) ranges from \( 0 \) to \( 2\pi \).

### Step 3: Set up the integral

In cylindrical coordinates, the volume element \( dV \) becomes \( r \, dx \, dr \, d\theta \). The triple integral becomes:

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

### Step 4: Evaluate the integral

1. **Integrate with respect to \( x \)**:

\[
\int_{2r^2}^2 7x \, dx = \left[ \frac{7x^2}{2} \right]_{2r^2}^2 = \frac{7(2^2)}{2} - \frac{7(2r^2)^2}{2} = \frac{28}{2} - \frac{7(4r^4)}{2} = 14 - 14r^4
\]

2. **Integrate with respect to \( r \)**:

\[
\int_0^1 (14 - 14r^4) r \, dr = 14 \int_0^1 r \, dr - 14 \int_0^1 r^5 \, dr
\]

These integrals are straightforward:

\[
14 \int_0^1 r \, dr = 14 \cdot \frac{r^2}{2} \Big|_0^1 = 14 \cdot \frac{1}{2} = 7
\]
\[
14 \int_0^1 r^5 \, dr = 14 \cdot \frac{r^6}{6} \Big|_0^1 = 14 \cdot \frac{1}{6} = \frac{14}{6} = \frac{7}{3}
\]

Thus, the result of integrating with respect to \( r \) is:

\[
7 - \frac{7}{3} = \frac{21}{3} - \frac{7}{3} = \frac{14}{3}
\]

3. **Integrate with respect to \( \theta \)**:

\[
\int_0^{2\pi} \frac{14}{3} \, d\theta = \frac{14}{3} \cdot 2\pi = \frac{28\pi}{3}
\]

### Final answer:
The value of the triple integral is \( \frac{28\pi}{3} \).

Would you like more details on any part of the solution?

Here are 5 related questions:
1. How would the limits change if the plane was \( x = 4 \)?
2. Can the integral be solved in Cartesian coordinates?
3. What if the paraboloid was \( x = 4y^2 + 4z^2 \)?
4. How does the symmetry of the region simplify the integration process?
5. What happens if the integrand is \( x^2 \) instead of \( 7x \)?

**Tip**: Converting to cylindrical coordinates simplifies integration for regions with circular symmetry.

Evaluate the triple integral 7x dV, where E is bounded by the paraboloid x = 2y^2 + 2z^2 and the plane x = 2.

Discover how to evaluate the triple integral ∫∫∫ 7x dV with boundaries defined by a paraboloid x = 2y^2 + 2z^2 and the plane x = 2. This step-by-step solution covers cylindrical coordinates, triple integrals, and the change of variables theorem, providing a clear understanding for students studying multivariable calculus.

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