part C please (a) Determine the surface defined by the parametric equations x = a sin(u) cos(v), y = b sin(u) sin(v), z = cs(u), 0 sus a, Osvs 2n.O hyperboloid of one sheetO elliptic paraboloidO ellipsoidO paraboloidO hyperboloid of two sheets(b) Use the parametric equations in part (a) to graph the surface for the case a = 3, b = 4, c = 3.10- 10106.y-10-5(c) Set up, but do not evaluate, a double integral for the surface area of the surface in part (b).du dvNeed Help?Watch ItRead It

Given:-x=a sin u cos v y=b sin u sin v z=c cos u0≤u≤π0≤v≤2πFormula Used:-For r=x(u, v)i→+y(u, v)j→+z(u, v)k→Surface Area is given by:-A=∬∥ru×rv∥dAx=a sin u cos vy=b sin u sin vz=c cos uHence r=a sin u cos vi→+b sin u sin v j→+ c cos u k→ ru=acos u cos vi→+bcos u sin vj→-c sin u k→ rv=-asin u sin vi→+bsin u cos vj→+0k→We now evaluate ru×rv ru×rv=i→j→k→acos u cos vbcos u sin v-c sin u-asin u sin vbsin u cos v0 =(cbsin2 u cos v)i→-j→(-ac sin2 u sin v)+k→(ab sin u cos u cos2 v+absin u cos usin2v) =(cbsin2 u cos v)i→+j→(ac sin2 u sin v)+k→(ab sin u cos u)|| ru×rv||=(cbsin2 u cos v)2+(ac sin2 u sin v)2+(ab sin u cos u)2 =c2b2 sin4 u cos2 v+a2c2 sin4 u sin2 v+a2b2 sin2 u cos2 u =sin uc2b2 sin2 u cos2 v+a2c2 sin2 u sin2 v+a2b2 cos2 u(c) Integral for the Surface area is A=∬∥ru×rv∥dA =∫02π∫0πsin uc2b2 sin2 u cos2 v+a2c2 sin2 u sin2 v+a2b2 cos2 u dudvis the required double integral for the surface area of the surfaceAs we want it for part(b)That is a=3, b=4, c=3Substituting,A=∫02π∫0πsin uc2b2 sin2 u cos2 v+a2c2 sin2 u sin2 v+a2b2 cos2 u dudv =∫02π∫0πsin u144 sin2 u cos2 v+81 sin2 u sin2 v+ 144cos2 u dudvANSWER:-∫02π∫0πsin u144 sin2 u cos2 v+81 sin2 u sin2 v+ 144cos2 u dudv