91影视

$$f(x, y,z) = z^{3} + z(x^{2} + 2^{y})$$ on the portion of the unit sphere that lies in the first octant, above the rectangle $$[0,\frac{1}{2}] \times [0,\frac{1}{3}]$$ in the xy-plane.

The equation of sphere is $$x^{2}+y^{2}+z^{2}=1$$

Here, we have $$z^{2}=1-x^{2}-y^{2}$$

$$\implies z^{3}=(1-x^{2}-y^{2})^{\frac{3}{2}}$$

Also, we have, $$z= \sqrt{1-x^{2}-y^{2}}$$

And, we find that, $$ds=\frac{1}{\sqrt{1-x^{2}-y^{2}}}dxdy$$

Now, evaluating the integral of the given function along the surface, we get

$$\int_{S} f(x,y,z)ds = \int_{0}^{\frac{1}{3}} \int_{0}^{\frac{1}{2}}(1-x^{2}-y^{2})^{\frac{3}{2}}+ ( \sqrt{1-x^{2}-y^{2}})(x^{2} + 2^{y})\frac{1}{\sqrt{1-x^{2}-y^{2}}}dxdy$$

Solving, we get

$$\int_{S} f(x,y,z)ds = \frac{61}{324}+\frac{2^{\frac{-2}{3}}}{ln2}-\frac{1}{2ln 2}$$