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With the help of gauss theorem find the stream of the vector field W=[x^2/9.y^2/9,z^2/9] through the part of the cone surface x^2+y^2=2z closed by the deck z=3.
 2 years ago
 2 years ago
With the help of gauss theorem find the stream of the vector field W=[x^2/9.y^2/9,z^2/9] through the part of the cone surface x^2+y^2=2z closed by the deck z=3.
 2 years ago
 2 years ago

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krobe8Best ResponseYou've already chosen the best response.1
Hi Raneen ... Here is how I see it, but I may be wrong. For clarity, I'm going to write "dot" for inner product of two vectors in formulas below. Gauss' theorem can be thought of in terms of fluid flow for an incompressible fluid. If you have a closed volume (the conic surface with deck at z=3), the flow of fluid into that volume, equals the flow out thru the boundary (surface) of the volume. Writing "div" for divergence, and V for volume (the portion of cone), and S for surface, Gauss' theorem is integral over V of (div dot W) d(volume) = integral over S of (W dot d(area)) In this particular problem, you are being asked to calculate the integral, over the surface, of W dot d(area). That is a nuisance of a problem, as you have to figure out the orientation of a patch of area  a bit of trig, not impossible but not really easy either. On the other hand, you can use Gauss' theorem. Form div dot W, and then integrate that over the volume. div dot W is partial W/ partial x + partial W/partial y + partial W/partial z which equals 2/9 times (x+y+z). It is a scalar. You then integrate that over the volume of the conic section. Probably using horizontal slices (perpendicular to z axis), but you can choose. And that's the answer! Best wishes, Ken R.
 2 years ago

raheenBest ResponseYou've already chosen the best response.0
thank you. but can you complete calculations because I's confused at this point.
 2 years ago

krobe8Best ResponseYou've already chosen the best response.1
I won't do the problem for you, but I'll walk you thru it. At some point you'll see what is going on, and race ahead to the conclusion, on your own. First, maybe it's good to visualize the volume. The surface is the points x^2 + y^2 = 2z with a "deck" z=3 placed on top. So, suppose y = 0 and draw, on xz axes, with x horizontal and z vertical, the equation x^2 = 2z. What does that look like?
 2 years ago

raheenBest ResponseYou've already chosen the best response.0
I think the whole problem is a cone shape
 2 years ago

raheenBest ResponseYou've already chosen the best response.0
it is a part of paraboloid?
 2 years ago

krobe8Best ResponseYou've already chosen the best response.1
Hi Raheen ... yes I was away. Right, it is a paraboloid. Looks like a parabola on paper 2dimensional projection: y=0, so equation is z = 1/2*x^2. Since x and y enter into the 3d equation as x^2 + y^2, for each fixed z (horizontal level) the slilce of the surface is a circle. Agree? I'm going away again. Back in a few hours. Bedtime here! Ken
 2 years ago

raheenBest ResponseYou've already chosen the best response.0
yes I agree, with the fixed z in space, we find any slice of the surface is just a circle. sweet dreams.
 2 years ago

krobe8Best ResponseYou've already chosen the best response.1
Ok then, suppose restrict to just one slice of volume, of thickness dz at position z, where z lies between 0 and 3. It is a circular slice, of radius sqrt(z/2). Imagine just the quadrant where x>0 and y>0. Integrate (2/9) * (x + y + z) over that quadrant. The z value is constant. The area is a double integral, integrated dx and dy, or you can do some polar coordinates version. But suppose just dx and dy. The integral breaks down into three integrals, one int(x dx dy), one int(y dx dy) and one int(z dx dy). The last one is just z (constant) times the area of quadrant. The first two are identical with x and y playing interchangeable roles. The only tricky part is getting the limits of integration right. But you can do that. Try a trig substitution. That's the way I see the problem. I don't actually know what will be the result, and am going to leave it to you to do the details of the integrations. Perhaps I've mistaken some aspect of this, but you (or some other reader of these posts) will likely correct any error I've made. The important point, going back to the original formulation of the problem, is that you've been able to evaluate integral of W dot d(area), which could be a nuisance, by using Gauss's theorem to instead evaluate "div W) d(volume). You can also do the earlier surface integral, W dot d(area), as a check, if you like. I'm sure there is more insight available by studying this problem. But the above is prettly much a good start, I believe. Best wishes, Ken
 2 years ago

raheenBest ResponseYou've already chosen the best response.0
thanks for you so much. that's so awesome
 2 years ago
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