To solve $f(x, y) = \sqrt{16 - x^2 - 16y^2}$ using the slice method, we'll look at slices of the graph in specific planes, particularly fixing either $x$ or $y$ and analyzing how the graph behaves.

Step-by-Step Solution:

The equation can be rewritten as: $f(x, y) = \sqrt{16 - x^2 - 16y^2}$ This is the upper half of an elliptical surface, since we are taking the square root, which results in non-negative $z$-values.

1. Slice along $x = 0$:

When $x = 0$, the equation simplifies to: $f(0, y) = \sqrt{16 - 16y^2} = 4\sqrt{1 - y^2}$ This describes a semicircle in the $yz$-plane, with a radius of 1 for $y$. The graph at $x = 0$ is a semicircle of radius 1.

2. Slice along $y = 0$:

When $y = 0$, the equation simplifies to: $f(x, 0) = \sqrt{16 - x^2}$ This describes a semicircle in the $xz$-plane, with a radius of 4. The graph at $y = 0$ is a semicircle of radius 4.

3. Slice along constant $y$ values:

For general constant values of $y = c$, the equation becomes: $f(x, c) = \sqrt{16 - x^2 - 16c^2}$ This is again a semicircle in the $xz$-plane, but with a decreasing radius. The radius is $\sqrt{16 - 16c^2} = 4\sqrt{1 - c^2}$, which means the radius of the semicircle depends on $c$, shrinking as $|c|$ increases from 0 to 1. When $|y| = 1$, the graph collapses to a point.

4. Graph Description in Words:

The surface described by $f(x, y)$ is an elliptical dome. At $x = 0$, the cross-section is a semicircle of radius 1. As you move along the $x$-axis, the cross-section becomes a semicircle with a radius of 4 when $y = 0$. The graph smoothly transitions, shrinking along the $y$-axis until it converges to a point at $y = \pm 1$.

In 3D, the graph is an ellipsoid with the following key properties:

• It stretches farther along the $x$-axis, with a maximum radius of 4.
• Along the $y$-axis, the ellipsoid has a radius of 1.

Would you like any specific details or clarifications on this? Below are a few related questions for further understanding:

5 Related Questions:

1. How do you derive the equation of an ellipsoid from a given equation?
2. What happens to the graph if we take the negative square root of the equation?
3. How would the graph change if the coefficient of $y^2$ was different?
4. What is the volume enclosed by this surface?
5. How would the graph look if we didn't take the square root (i.e., used $z^2 = 16 - x^2 - 16y^2$)?

Tip:

When using the slice method, it is helpful to visualize each cross-section of the surface in 2D and then mentally combine them to build the full 3D shape.