91Ó°ÊÓ

The function:

$f(x,y,z)=e^z$

The formula for finding the integral over the smooth surface is given by,

${∫}_{S}dS=∫{∫}_{D}f\left(x\right(u,v),y(u,v),z(u,v\left)\right)||r_u×r_v||dudv$

The surface is parametrized as follows:

$$\begin{gathered} r(u,v)=(\sin u\cos v,\sin u\sin v,\cos u) \\ {r}_u=\frac{∂r}{∂u} \\ =(\cos u\cos v,\cos u\sin v,-\sin u) \\ {r}_v=\frac{∂r}{∂v} \\ =(-\sin u\sin v,\sin u\cos v,0) \end{gathered}$$

$$\begin{gathered} r_u×r_v=(\cos u\cos v,\cos u\sin v,-\sin u)×(-\sin u\sin v,\sin u\cos v,0) \\ ={\sin }^{2}u\cos vi+{\sin }^{2}u\sin vj+\sin u\cos uk \\ ||r_u×r_v||=\sqrt{{({\sin }^{2}u\cos v)}^2+{({\sin }^{2}u\sin v)}^2+{(\sin u\cos u)}^2} \\ =\sqrt{{\sin }^{4}u+{\sin }^{2}u{\cos }^{2}u} \\ =\sqrt{{\sin }^{2}u({\sin }^{2}u+{\cos }^{2}u)} \\ =\sin u \end{gathered}$$

$$\begin{gathered} f(x,y,z)=e^z \\ f\left(x\right(u,v),y(u,v),z(u,v\left)\right)=e^{\cos u} \end{gathered}$$

The region of integration is given by,

$$\begin{gathered} D=\left\{\right(u,v\left)\overline{)0≤u≤\frac{\mathrm{π}}{2},0≤v≤\frac{\mathrm{π}}{2}}\right\} \\ {∫}_{S}dS=∫{∫}_{D}f\left(x\right(u,v),y(u,v),z(u,v\left)\right)||r_u×r_v||dudv \\ ={∫}_0^{\frac{\mathrm{π}}{2}}{∫}_0^{\frac{\mathrm{π}}{2}}{e}^{\cos u}\sin ududv \\ ={∫}_0^{\frac{\mathrm{π}}{2}}{[-e^{\cos u}]}_0^{\frac{\mathrm{π}}{2}}dv \\ ={∫}_0^{\frac{\mathrm{π}}{2}}[-e^0+e^1]dv \\ ={∫}_0^{\frac{\mathrm{π}}{2}}(e-1)dv \\ =(e-1){\left[v\right]}_0^{\frac{\mathrm{π}}{2}} \\ =\frac{\mathrm{π}}{2}(e-1) \end{gathered}$$