Find the volume of a solid bounded above by the function f(x,y) = 2x2 + 4y2 +1, below by the xy-plane, for 4 < x < 8 and -2 < y < 11. Leave your final answer correct to 2 decimal places.

To find the volume of the solid bounded above by the function \( f(x, y) = 2x^2 + 4y^2 + 1 \), and below by the xy-plane, over the region defined by \( 4 < x < 8 \) and \( -2 < y < 11 \), we can set up and evaluate the double integral:

\[
V = \int_{y=-2}^{11} \int_{x=4}^{8} f(x, y) \, dx \, dy
\]

This can be written explicitly as:

\[
V = \int_{y=-2}^{11} \int_{x=4}^{8} (2x^2 + 4y^2 + 1) \, dx \, dy
\]

### Step 1: Integrate with respect to \(x\)

\[
\int_{x=4}^{8} (2x^2 + 4y^2 + 1) \, dx
\]

This integral can be split as:

\[
\int_{x=4}^{8} 2x^2 \, dx + \int_{x=4}^{8} 4y^2 \, dx + \int_{x=4}^{8} 1 \, dx
\]

Calculating each term separately:

- For \( \int_{x=4}^{8} 2x^2 \, dx \):

\[
\int 2x^2 \, dx = \frac{2x^3}{3}
\]

Evaluate from 4 to 8:

\[
\left[ \frac{2(8)^3}{3} - \frac{2(4)^3}{3} \right] = \frac{2(512)}{3} - \frac{2(64)}{3} = \frac{1024}{3} - \frac{128}{3} = \frac{896}{3}
\]

- For \( \int_{x=4}^{8} 4y^2 \, dx \):

\[
\int 4y^2 \, dx = 4y^2 \cdot x
\]

Evaluate from 4 to 8:

\[
4y^2 \cdot (8 - 4) = 16y^2
\]

- For \( \int_{x=4}^{8} 1 \, dx \):

\[
\int 1 \, dx = x
\]

Evaluate from 4 to 8:

\[
8 - 4 = 4
\]

Thus, the integral with respect to \( x \) becomes:

\[
\frac{896}{3} + 16y^2 + 4
\]

### Step 2: Integrate with respect to \(y\)

Now, integrate this result with respect to \( y \) over the interval \(-2\) to \(11\):

\[
\int_{y=-2}^{11} \left(\frac{896}{3} + 16y^2 + 4\right) \, dy
\]

This integral can be split as:

\[
\frac{896}{3} \int_{y=-2}^{11} 1 \, dy + 16 \int_{y=-2}^{11} y^2 \, dy + 4 \int_{y=-2}^{11} 1 \, dy
\]

Evaluating each integral:

- For \( \frac{896}{3} \int_{y=-2}^{11} 1 \, dy \):

\[
\frac{896}{3} \cdot (11 - (-2)) = \frac{896}{3} \cdot 13 = \frac{11648}{3}
\]

- For \( 16 \int_{y=-2}^{11} y^2 \, dy \):

\[
\int y^2 \, dy = \frac{y^3}{3}
\]

Evaluate from \(-2\) to \(11\):

\[
\left[\frac{11^3}{3} - \frac{(-2)^3}{3} \right] = \frac{1331}{3} - \frac{-8}{3} = \frac{1339}{3}
\]

Multiply by 16:

\[
16 \cdot \frac{1339}{3} = \frac{21424}{3}
\]

- For \( 4 \int_{y=-2}^{11} 1 \, dy \):

\[
4 \cdot (11 - (-2)) = 4 \cdot 13 = 52
\]

Adding all these up:

\[
V = \frac{11648}{3} + \frac{21424}{3} + 52
\]

Combine the fractions:

\[
V = \frac{11648 + 21424}{3} + 52 = \frac{33072}{3} + 52 = 11024 + 52 = 11076
\]

### Final Answer
The volume of the solid is \( \boxed{11076.00} \) cubic units, correct to two decimal places.

Would you like any more details or have any questions? Here are some related questions you might find interesting:

1. How do you set up a double integral over a non-rectangular region?
2. What changes if the function above the region is more complex?
3. How do you interpret the physical meaning of this volume in different contexts?
4. Can you extend this concept to triple integrals for volumes in 3D?
5. What is the significance of the bounds in determining the volume?

**Tip:** Breaking down complex integrals into simpler parts helps in evaluating them step by step accurately.

Learn how to calculate the volume of a solid bounded above by the function f(x, y) = 2x^2 + 4y^2 + 1 and below by the xy-plane over the region 4 < x < 8 and -2 < y < 11. Step-by-step solution provided.

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