Find the surface area of a portion of a sphere x^2 + y^2 + z^2 = 9z that lies inside the paraboloid x^2+y^2=4z

Question

1st the spere is centered at z=4.5 and with radius 4.5 also
Right?

2nd the area is at the top of the sphere (inside the paraboloid)... but I can't find the limits of the area-- the projection of the area to the xy plane
that s for setting up the INT
look at James Stewart's book chap 15 & 16.

What I tried to do was subst. z=para- with z at sphere, but it was disastrous producing x^4 and y^4
right?

so I switch to spherical coordinate method of INT
here I tried to find the phi-Φ angle .. see stewart chapter 15..
and also the rho-ρ would change.. when we INT right ?

honestly I kinda suck at spherical coord there too..
I am still trying..

Actually I am quite pesimmistic that anyone here would be able to correctly solve it.. although the problem seem to be quite simple..
since for many times I have asked about problems from chap 15-16 stewart calculus..most of the time the ans were wrong..even thou they have really tried

So BIG thx and salute to the solver!!

or is it using vector r(u,v)
and then cross product etc.. then INT
****!?!?

Answer 1

To answer your first two questions, yes, and yes. After completing the square to get into a standard form, that's what you get. . . . Now, we COULD find y(x) on the plane of intersection to form the bounds for an integral, but that's stupid. Instead take the two equations and solve for z to find the plane that slices off the section of interest, e.g. z=5. The center of the circle is just below, at z=4.5. We want the area above z=5. Forgetting coordinates for the moment, let's just integrate rings to find the surface area of a sphere. Think of a hemisphere, and R goes up counterclockwise with theta, e.g. x=Rcos(theta) y= Rsin(theta) The area of a little slice is r*d(theta), the width of slice, * 2Pi*(Rcos(theta)), the circumference/length of the slice. For a hemisphere, integrate 2*pi*R^2*cos(theta) from theta=0 to theta=90 deg and not surprisingly you find A=2*pi*R^2 (half the surface area of a sphere). But we don't want the whole hemisphere, just that part above z=5, which is .5 above the center. So start with that theta=arcsin(.5/R) =6.4 degrees instead of 0. Then you'll find the area is 2PiR^2*( 1-sin(6.4deg))=113 square units.

Peace.

Answer 2

What branch of mathematics are you folks in? Very impressive, both the jargon and the actual knowledge, I don't remember any of this from HS math. Cool stuff.