Homework Statement
I'm just required to setup the integral for the question posted below
Homework EquationsThe Attempt at a Solution
So solving for phi @ the intersection of the sphere and the plane z=2:
z = pcos(phi)
2 = 3cos(phi)
phi = arccos(2/3)
so my limits for phi would go from 0 to...
oops made a sign error
Im getting cos(phi) +/- sqrt(1+cos^2(x) )
And btw is it wrong to use geometry to get rho and phi in certain cases?
For example this qn(Not switching the topic of the thread btw...just found a question where I think geometry may be applicable)
Sorry about that...didn't get to respond because of assignments etc for other courses.
But I worked that out already but I didnt post it so here it is:
x(sq)+y(sq) + z(sq)-2az+a(sq) = 2a(sq)
rho(sq)-2az = a(sq)
rho(sq)-2a(rho)(cos(phi)) = a(sq)
rho(sq)-2a(rho)(cos(phi)) -a(sq) = 0
Using the...
This was the diagram I used to get phi from the yz plane.
I just did cos(phi) which is 2a/rho and found rho from that to get 2asec(phi). But I understand where you're coming from...if I change the eqns I get z = 2r and z=sqrt(2a^2-r)+a.
But for spherical...I can't use the diagram like that to...
Ohh I see.
From there I graphed the region onto the yz plane.
So its the graph z= 2|y| intersected by the sphere. I then used a bit of trigs to get that cos(phi) = 2a/rho giving rho as 2asec(phi) and tan(phi) as 1/2 -> phi = arctan(1/2)So my bounds were:
0<rho<2asec(phi)
0<phi<arctan(1/2)
Well I decided to use spherical coordinates from the start and normally to get my bounds for rho and phi I set x=0 to get the projection on the yz plane. So from the cone's equation I ended up with z=2y and I substituted z=2a into that which gave y=a.
Hmm okay so I solved for z already which I got as 2a ---> Resulting in y = a. Substituting those values back into the equation for the cone I get x = 0.
So the curves intersect in the yz plane?
But if you substitute z=2a in that same eqn for the cone you get a circle of radius a. So the region...
lol I thought it would have been easier to understand straight from the textbook.
But it said z>=0 so at 0 wouldn't the solid be lying on the x-y plane?
Because if I substitute z= 0 into the equation of the sphere then you get a circle
Hey guys I've been working some triple integration problems and I've stumbled across a question that I'm having problems with
So from the picture below my solution is incorrect and I can't seem to figure out where I went wrong. Is my setup for the integrals correct or is that where I've made my...
Ahh yes they do...the parabola closes down on the xy plane. That means I will be integrating f(x,y) over the region bounded by the two sine functions and the lines y=2 and y = -2 with the order of integration being dxdy
Sorry about the bolding...didn't know...just used the default layout.
Thanks btw. I slept over it and I was thinking to use the z=f(x,y) function as my integrating function as you said over the region bounded by the two curves.
But why use between y=2 & y=-2 though? I know that I have to have...