Respuesta :
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
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 [tex]x^{2} -y^{2}+z^{2}=2[/tex]
1. Slice for y = 2.
We replace this value in the given volume.
[tex]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[/tex]
So, results in a circumference with radius [tex]\sqrt{6}[/tex], because a circumference is defind as [tex]x^{2} +y^{2} =r^{2}[/tex]. (On plane XY).
2. Slice for x = 1.
We repeat the process.
[tex](1)^{2} -y^{2}+z^{2}=2\\z^{2}-y^{2}=2-1\\z^{2}-y^{2}=1[/tex]
It forms a horizontal hyperbola on plane ZY.
3. Slice for y = 0.
[tex]x^{2} -0^{2}+z^{2}=2\\x^{2}+z^{2}=2[/tex]
Another circle with radius of [tex]\sqrt{2}[/tex] on plane XZ.
4. Slice for x = 2.
[tex](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[/tex]
It forms a hyporbola on plane YZ.