Question

S:x^2+y^2+z^2-2x-4y-4x=0, π:2x+y+2z-5=0
find the circle area of sphere S cut by plane π.

Answer 1

-2x-4y-4x=-4y-6x
sure that's not a typo? why is it unsimplified?

Answer 2

-2x-4y-4z
maybe?

Answer 3

i don't know the answer. not typo

Answer 4

the answer must be circle i think.

Answer 5

S:x^2+y^2+z^2-2x-4y-4x=0
^ ^
are you *sure* these are supposed to both be x's ?

Answer 6

it's just weird to have a problem like that, where the first step is algebra I level simplification...

Answer 7

S is sphere, and pi is plane. when the plane cut the the sphere we get a circle. is it true to think like this

Answer 8

ok well if you insist that there is no typo, yes it is a sphere
we can complete the square in x and y to get it into standard form

Answer 9

hm... I have to think about getting the intersect a bit...

Answer 10

I graphed the equations.. still dunno how

Answer 11

we gotta eliminate one of the variables somehow with a substitution from the plane equation into the sphere one

Answer 12

hey that is a z...
you said there were no typos :/

Answer 13

you wrote
x^2+y^2+z^2-2x-4y-4x
above and
x^2+y^2+z^2-2x-4y-4z
on your link

Answer 14

ok, since we have\[2x+y+2z-5=0\]we also have that\[10-4x-2y-4z=0\](multiplication by -2)

since this and the sphere equation are both equal to zero we can set them equal to each other\[10-4x-2y-\cancel{4z}=x^2+y^2+z^2-2x-4y-\cancel{4z}\]we can cancel a bit\[x^2+y^2+z^2+2x-2y-10=0\]now solve for z in the plane equation and substitute that in...

Answer 15

\[z=\frac52-x-\frac12y\]\[x^2+y^2+(\frac52-x-y)^2+2x-2y-10=0\]hm... now I am thinking that we won't be able to complete the square, which was what I wanted to do here...

Answer 16

ok I just read an interesting approach from wikipedia
let me try it out!
http://en.wikipedia.org/wiki/Plane%E2%80%93sphere_intersection

Answer 17

ok ill try too

Answer 18

first we need the radius of the sphere, which we can find by completing the square (or with software of course)

Answer 19

do you want to see how to get it by completing the square?

Answer 20

I'll just imagine that you know it I guess...
then you need the equation of the plane in the form\[ax+by+cz=d\]such that\[a^2+b^2+c^2=1\]since for yours we get\[2x+y+2z=5\]we need to change the coefficients, which we do my dividing by the magnitude of the gradient of the plane

Answer 21

\[\|\nabla f\|=\sqrt{2^2+1^2+2^2}=3\]so we get\[\frac23x+\frac13y+\frac23z=\frac53=d\]and the radius of the sphere is \[r=3\]

Answer 22

according to the article, it says of the intersection of these forms
"if d < r, (it forms) a circle of intersection, with center (da, db, dc) and squared radius r^2 − d^2."

Answer 23

so apparently the square of the radius of that circle that is formed by the intersection should be\[R^2=r^2-d^2=3^2-(\frac53)^2=\frac{81}9-\frac{25}9=\frac{56}9\]hence \[A=\pi R^2=\frac{56\pi}9\]

Answer 24

at lest that's what I get from the wiki article...

Answer 25

least*

Answer 26

you really help a lot.

Answer 27

thanks, I really learned a lot :)

Answer 28

*

Answer 29

what's that mean