Evaluate the surface integral s f · ds for the given vector field f and the oriented surface s. in other words, find the flux of f across s. for closed surfaces, use the positive (outward) orientation. f(x, y, z) = xz i + x j + y k s is the hemisphere x2 + y2 + z2 = 81, y ≥ 0, oriented in the direction of the positive y-axis

Because I've gone ahead with trying to parameterize directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over straight away, let's close off the hemisphere with the disk of radius 9 centered at the origin and coincident with the plane . Then by the divergence theorem, since the region is closed, we have

where is the interior of . has divergence

so the flux over the closed region is

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

Parameterize by

with and . Take the normal vector to to be

Then the flux of across is

the sum of all the angles in a triangle are equal to 180

so you can add   the 3 given angles together to equal 180:

5y + 2y + 2y = 180

or since all three angles have the same letter, combine the 3 angles to equal 180:

9y = 180

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.