Spheres and Cylindrical Surfaces

Topics:

• Know the standard equation of a sphere and be able to find the center and radius of a sphere.
• Be able to sketch cylindrical surfaces.

so much for cylindrical surfaces. we covered these in class but idk

In 2 dimensions $D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$. Thus, in 3D the distance between $P(a,b,c)\text{ and }Q(x,yz)$ is given as $D=\sqrt{(x-a{)}^2+(y-b{)}^2+(z-c{)}^2}$

If we’re taking about a sphere, the standard equation for a sphere is given as $r^2=(x-a{)}^2+(y-b{)}^2+(z-c{)}^2$, this should look familiar to the equation for a circle

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Example Find an equation of the sphere with the enter is $P(4,-6,0)$ and a diameter of $8$

$r=\frac{8}{2}=4$

$(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$

$(x-4{)}^2+(y+6{)}^2+z^2=16$

Example Find an equation of the sphere which contains the points $P(8,-1,9)$ and $Q(6,-9,-1)$ and has a diameter equal to the distance between $P$ and $Q$

$D(P,Q)=\sqrt{(8-6{)}^2+(-1-(-9){)}^2+(9-(-1){)}^2}$ $D(P,Q)=\sqrt{2^2+8^2+10^2}$ $D(P,Q)=\sqrt{168}=2\sqrt{42}$ $\text{Thus, }r=\sqrt{42}$

Now, we find our midpoint of $\overline{\mathrm{PQ}}$ , we can use a modified version of the distance formula! $(\frac{8+6}{2},\frac{-1+(-9)}{2},\frac{9+(-1)}{2})$ $(7,-5,4)$

$\sqrt{(x-7{)}^2+(y+5{)}^2+(z-4{)}^2}=\sqrt{42}$

$(x-7{)}^2+(y+5{)}^2+(z-4{)}^2=42$

Example Describe all the points in space whose coordinates satisfy the following inequality: $x^2+y^2+z^2-8y-2z+16>0$

$(x^2)+(y^2-8y)+(z^2-2z)+(16)>0$

idk just complete the square this is left as an exercise to the reader :)

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1/8/25