Question

Verify Stokes theorem for F =(y^2 + x^2 - x^2)i + (z^2 + x^2 - y^2)j + (x^2 + y^2 - z^2)k over the portion of the surface x^2 + y^2 -2ax + az = 0

Answer 1

Given lean is x²ty²-29x+ az zo The equation of the given surface can be written as (x-aj? + y2 = -a (Z-a) This shows that the given surface is a paraboloid of revolution with its vedtex at (a,o,a) and it meets the plane 70 in the circle a given by (x-2)! + y2 = 0,7=0 In parametra coordinates, any point on the circumference of the above cirde o is given by 2= a +a coso, y=a sino, 220 Along this circe we're F = (y²t2²-x² i + (2²+x²-y², j tox²+ y²-2²) k = { a² sinho - a² cit coso ²h î tha² citroso? -q sino} + {a² citrose ² a²sin ? Ĉ dă= idx+ jdy +kd? = (dx ; + dy jtdze k)d = (-asino i tacosoj) do HfF.dă = a? pia [- sino -ci trosoj? sino + { cit coses-sinio) osald selfs tot hot fciado? _timo, do z le coso s circose)-inoldo (: sn (97-0)= =simo ] COS (25-0) =(050 203 cos o ltcosto + 21050-sino) do To 2x243 n acosado ('COSUT-Ow--coso ) - Folf" comodo = saya nota doo Now by Stoke's theorem, the surface integral over the surface of the paraboloid is the same as the surful integral overs the plane regions of the circle as

the same boundary c both s . also and curl s have F = 1 रिप- Ty?t?? x²+x2-y? x²+² 7² L = 2cy-z' î + 207-21 ſ + 2(x-ysk abo ñ for the circles S1 ) = K . The surfure integral over s= f courl & ñ ds, - acx-yids, ds, = rdo dr be an element of this circle where r,o are the polar coordinates of the element worto the center of the cirule then a= arracoso y=rsino ST. + roos - rsino) rdodr rzo 850 2t T* ( a + Tìne + cos8 ) ra. = 2.205 rdr = ara² (2) from (1) and (2) Stoke's Theorem is verified!