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.