Question

What is the outward flow of the vector field F=xi+xj+k from the part of the z=x^2-y^2 surface inside the x^2+y^2=a^2 in cylinder?

Answer 1

Ans :- field also defined as the Given to find the outward flow of the vector F=xitxj tk from the part 2=002 ya surface inside the x2 + y2 = as in the cylinder. The outward flow is flux of the surface integral of the vector field. Outward flow s Fonds S where i is the normal unit vector. Outward flow = lafliege (xitxi tk) (xxi txj tk) doody -T1-ya a Outward flow Jioya (x2 + x2 + 1) dx dy -Ji-ya Outward How= linya ex*+ de dy Si-ya Where -a to a. og limits are X limits are - Ji-42 to 51-42. Therefore, 119.92 a 2x2 ti doc dy. outward flow = -a - 11-42

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