Homework Statement
Use the Divergence Theorem to evaluate ∫∫S (8x + 10y + z2)dS where S is the sphere x2 + y2 + z2 = 1.
Homework Equations
∫∫S F dS = ∫∫∫B Div(F) dV
The Attempt at a Solution
I dunno, this isn't a vector field so I don't know how to take the divergence of it so...
Homework Statement
Let r = x i + y j + z k and R = |r|. Let F = r/R^p.
find div(F) in terms of r.. i can't figure out how to express it in therms of r
Homework Equations
div(F) = the gradient added together
Separate the top of the original function, cancels out an i.
Then make a common denominator and add the top of both functions.
z= 2/i(-3+4i) + i/i(-3+4i)
z= 2/(-3i-4) + 1/(-3+4i)
z= 2(-3+4i) / (-3i-4)(-3+4i) + 1(-3i-4)/(-3i-4)(-3-4i)
etc..
so i have the function z=(2+i)/(i(-3+4i)) and i need to linearize it to find the Im(z) and Re(z)
I get down to z= (-6 +8i -3i -4 )/ (9i +12 +12 -16i) which i then simplify down to
z= (5i -10)/(-5i+24)
However when solve it i get a different answer from wolfram (from when i plugged...
Yaaa, i figured it out.
Just multiplying both sides of the r function by r, then subbing in y=rsin(theta) and x=rcos(theta) then using sin^2 + cos^2 =1 will make it just in terms of x and y.
Homework Statement
x=2acos(theta- pi/3)cos(theta)
y=2acos(theta- pi/3)sin(theta)
Write everything in terms of x and y
Homework Equations
cos(a-b)=cos(a)cos(b)+sin(a)sin(b)
sina/cosa=tana
etc...
The Attempt at a Solution
I've tried a lot of rearranging, can't figure it out.
Okay,
I'm curious though why you wrote the numerator with a 1 in it. Since it is n+1 won't it start at 1/2 and finish with 1/n+1, and the denominator start as 1 and finish with 1/n?
Probably try trig sub, with t equal to sec(theta)
don't forget to tack on a sec(theta)tan(theta) when you change dt to dtheta
Edit: I meant sec(theta)tan(theta) not tan^2(theta)
so it becomes
Integral of sec(theta)tan^{2}(theta) dtheta
Okay, makes sense.
honestly though I don't quite understand the algebra of that limit
n+ 1 / n
you would have, (1/2 + 1/3 ... 1/n+1 ) / (1+1/2... 1/n)
Everything cancels out but ( 1/ n+1 ) / 1 ) ?
Or does the 1/n stay on the bottom
You could try the washer method but it wouldn't really make sense because if you look at the graph, taking horizontal slices would mean that you would have to change your limits of integration and solve for two integrals. One from y=0 to y=1/5 and the radius would just be constant, and the other...
Hmm,
But how would showing the sum in brackets goes to one with the ratio test help? If that limit equals one the ratio test would be inconclusive.
If I treat this as two sums the (1/n) sum would diverge. So I still don't know what to do.
Hmm, I think you just do the sum of a sum..
But if you do that and use the ratio test you get the limit to be zero, and then it would be convergent for all values of x. I just have no clue if that's how it's done.