Question

Consider the level surface given by

Answer 1

Answer:

First of all, each slice represents an intersecting plane at that level.

For example, y = 2 is a plane that passes thorugh that level and cuts the volume given by $x^{2} -y^{2}+z^{2}=2$

1. Slice for y = 2.

We replace this value in the given volume.

$x^{2} -y^{2}+z^{2}=2\\x^{2} -(2)^{2}+z^{2}=2\\x^{2}+z^{2}=2+4\\x^{2}+z^{2}=6$

So, results in a circumference with radius $\sqrt{6}$, because a circumference is defind as $x^{2} +y^{2} =r^{2}$. (On plane XY).

2. Slice for x = 1.

We repeat the process.

$(1)^{2} -y^{2}+z^{2}=2\\z^{2}-y^{2}=2-1\\z^{2}-y^{2}=1$

It forms a horizontal hyperbola on plane ZY.

3. Slice for y = 0.

$x^{2} -0^{2}+z^{2}=2\\x^{2}+z^{2}=2$

Another circle with radius of $\sqrt{2}$ on plane XZ.

4. Slice for x = 2.

$(2)^{2} -y^{2}+z^{2}=2\\z^{2}-y^{2}=2-4\\z^{2}-y^{2}=-2\\\frac{z^{2}-y^{2}}{-2} =\frac{-2}{-2} \\\frac{y^{2} }{2} -\frac{z^{2} }{2} =1$

It forms a hyporbola on plane YZ.

Answer 2

1. circle of radius sqrt(6) in x-z plane 2. hyperbola in y-z plane opening in z direction3. circle of radius sqrt(2) in x-z plane 4. hyperbola in y-z plane opening in y direction