what is the volue integral of -1 for the surface z = 4-4x^2-y^2 for z from 0 to 4

To find the volume integral of \(-1\) over the surface defined by \( z = 4 - 4x^2 - y^2 \) for \( z \) from \( 0 \) to \( 4 \), we need to set up and evaluate the appropriate volume integral.

### Step 1: Understanding the surface

The surface equation \( z = 4 - 4x^2 - y^2 \) is a paraboloid, and we are asked to find the volume between this surface and the plane \( z = 0 \), where the volume element is multiplied by \(-1\).

### Step 2: Setting up the volume integral

In Cartesian coordinates, the volume of a region is typically calculated as:

\[
V = \iiint_{\text{region}} 1 \, dV
\]

However, in this case, we are asked to integrate \(-1\) over the volume, so the integral becomes:

\[
V = \iiint_{\text{region}} -1 \, dV
\]

Since the surface is symmetric and describes a paraboloid, it is convenient to switch to cylindrical coordinates where the paraboloid equation becomes:

\[
z = 4 - 4r^2
\]

where \( r^2 = x^2 + y^2 \), and the volume element in cylindrical coordinates is \( dV = r \, dr \, d\theta \, dz \).

### Step 3: Limits of integration

The limits for \( z \) are from \( 0 \) to \( 4 \), and we need to determine the corresponding range for \( r \) at \( z = 0 \) and \( z = 4 \).

- When \( z = 4 \), \( r = 0 \).
- When \( z = 0 \), solving \( 0 = 4 - 4r^2 \) gives \( r = 1 \).

Thus, \( r \) ranges from \( 0 \) to \( 1 \), and \( \theta \) ranges from \( 0 \) to \( 2\pi \).

### Step 4: The volume integral

The integral becomes:

\[
V = \int_0^{2\pi} \int_0^1 \int_0^{4 - 4r^2} -1 \, r \, dz \, dr \, d\theta
\]

First, integrate with respect to \( z \):

\[
\int_0^{4 - 4r^2} -1 \, dz = -(4 - 4r^2)
\]

Substitute this into the integral:

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

### Step 5: Solving the integral

Now we solve the remaining integrals:

\[
V = \int_0^{2\pi} \int_0^1 -(4r - 4r^3) \, dr \, d\theta
\]

First, integrate with respect to \( r \):

\[
\int_0^1 (4r - 4r^3) \, dr = \left[ 2r^2 - r^4 \right]_0^1 = 2(1) - (1) = 1
\]

Thus, the integral becomes:

\[
V = \int_0^{2\pi} -1 \, d\theta = -2\pi
\]

### Step 6: Final result

The volume integral of \(-1\) over the given surface is:

\[
V = -2\pi
\]

Let me know if you'd like more details or if you have any questions!

Here are some related questions you might explore:
1. How does changing the limits of \( z \) affect the volume integral?
2. How would the integral change if the surface was defined in spherical coordinates?
3. What if the surface was a different paraboloid, say \( z = 5 - 3x^2 - 2y^2 \)?
4. How would the result change if we integrated a different function over the same region?
5. How does the symmetry of the surface simplify the integration process?

**Tip:** Converting to cylindrical or spherical coordinates often simplifies integration when dealing with symmetric regions.

what is the volume integral of -1 for the surface z = 4-4x^2-y^2 for z from 0 to 4

This detailed solution shows how to compute the volume integral of -1 over the surface defined by the paraboloid z = 4 - 4x^2 - y^2, with limits of z from 0 to 4. It covers the use of cylindrical coordinates and integration techniques, suitable for students studying multivariable calculus.

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