Answer:-
a)
b)
z=x2+y2
zx=2x,zy=2y
Change in polar coordinates
x=r Cosθ\thetaθ
y=r Sinθ\thetaθ
x2+y2=r2
dA=rdrdθ\thetaθ
1<r<√3,0<θ\thetaθ <2θ\thetaθ
(1+zx2+zy2)=(1+4x2+4y2)\sqrt (1+z_x^{2}+z_y^{2})=\sqrt(1+4x^{2}+4y^{2})(1+zx2+zy2)=(1+4x2+4y2)
=(1+4r2)\sqrt(1+4r^{2})(1+4r2)
Surface area=∬R(1+zx2+zy2)dA\iint_R\sqrt(1+z_x^{2}+z_y^{2})dA∬R(1+zx2+zy2)dA
=∫02π∫1√3(1+4r2)rdrdθ\int_{0}^{2π}\int_{1}^{√3}\sqrt(1+4r^{2})rdrd\theta∫02π∫1√3(1+4r2)rdrdθ
=∫02πdθ∫1√3(1+4r2)rdr\int_{0}^{2π}d\theta\int_{1}^{√3}\sqrt(1+4r^{2})rdr∫02πdθ∫1√3(1+4r2)rdr
Let 1+4r2=t2
8rdr=2tdt
∴\therefore∴ (2π-0)∫√5√13t∗tdt4\int_{√5}^{√13}t*\frac{tdt}{4}∫√5√13t∗4tdt
=2π4[t33]√5√13\frac{2π}{4}[\frac{t^{3}}{3}]_{√5}^{√13}42π[3t3]√5√13
=π6(13√13−5√5)\frac{π}{6}(13√13-5√5)6π(13√13−5√5) square units
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