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.