if i want to evaluate a function and the bounds are defined by z=x^2+y^2 and z = 9 what would the 3 sets of bounds be?
thanks ahead of time!
2007-03-11 20:47:49 · 3 answers · asked by need help! 3 in Science & Mathematics ➔ Mathematics
the first equation is a cone, the second a plane. I want the bounds of the triple integral defined by the region contained... please?
2007-03-11 20:53:54 · update #1
it is not a sphere
2007-03-11 20:54:20 · update #2
Your bounds on z would be 0 and 9. The upper bound on x is where x^2=9 or x=3, same for y. At any value of z, the x-y section is a circle of radius √z. The radius at z = 9 is 3, and that is the extent of the volume at it's maximum z. Everything joins at x=y=z=0, so that is the lower bound for all axes.
2007-03-11 21:01:54 · answer #1 · answered by gp4rts 7 · 1⤊ 0⤋
not sure there is 3. the first equation gives you a sphere and the second gives a plane. so just picture a plane intersecting a sphere. you get a circle.
2007-03-12 03:51:50 · answer #2 · answered by brandon 5 · 0⤊ 1⤋
z=9; x=3cost and y=3cost it is a sphere with center (0,0,9)
2007-03-12 03:52:39 · answer #3 · answered by djin 2 · 0⤊ 1⤋