div F=x

a) The given surface \(\displaystyle{x}^{{2}}+{y}^{{2}}\leq{z}\leq{1}\) is the paraboloid \(\displaystyle{x}^{{2}}+{y}^{{2}}={z}\) under the plane z=1. In cylinderical co-ordinates the region bounded by the surface is given by: \(\displaystyle{E}={\left[{\left({r},\theta,{z}\right)}:{0}\leq\theta\leq{2}\pi,{0}\leq{r}\leq{1},{r}^{{2}}\leq{z}\leq{1}\right]}\)

By Gauss theorem, \(\displaystyle\int\int_{{S}}{F}\cdot{d}{s}=\int\int\int_{{E}}\div{F}{d}{V}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{1}}}{\int_{{{r}^{{2}}}}^{{1}}}{r}{\cos{\theta}}{r}\ {\left.{d}{z}\right.}{d}{r}{d}\theta\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{1}}}{r}^{{2}}{\cos{\theta}}{\left({1}-{r}^{{2}}\right)}{d}{r}{d}\theta\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\cos{\theta}}{d}\theta{\int_{{0}}^{{1}}}{\left({r}^{{2}}-{r}^{{4}}\right)}{d}{r}\)

\(\displaystyle{{\left[{\sin{\theta}}\right]}_{{0}}^{{{2}\pi}}}{{\left[{\frac{{{r}^{{3}}}}{{{3}}}}-{\frac{{{r}^{{5}}}}{{{5}}}}\right]}_{{0}}^{{t}}}={0}\)

b) Note that here we have an extra condition \(\displaystyle{x}\geq{0}\) which means that we have that portion of paraboloid under the plane which lies in front ofyz-plane. Then the region boundelby this surface is:

\(\displaystyle{E}={\left[{\left({r},\theta,{z}\right]}\right)}:-{\frac{{\pi}}{{{2}}}}\leq\theta\leq{\frac{{\pi}}{{{2}}}},{0}\leq{r}\leq{1},{r}^{{2}}\leq{z}\leq{1}{]}\)

\(\displaystyle\int\int{F}\cdot{d}{s}=\int\int\int_{{E}}\div{F}{d}{V}\)

Then

\(\displaystyle={\int_{{-{\frac{{\pi}}{{{2}}}}}}^{{\frac{{\pi}}{{{2}}}}}}{\int_{{0}}^{{1}}}{\int_{{{r}^{{2}}}}^{{1}}}{r}{\cos{\theta}}{r}{\left.{d}{z}\right.}{d}{r}{d}\theta\)

\(\displaystyle={{\left[{\sin{\theta}}\right]}_{{-{\frac{{\pi}}{{{2}}}}}}^{{{\frac{{\pi}}{{{2}}}}}}}{{\left[{\frac{{{r}^{{3}}}}{{{3}}}}-{\frac{{{r}^{{5}}}}{{{5}}}}\right]}_{{0}}^{{1}}}\)

\(\displaystyle={2}\times{\frac{{{2}}}{{{15}}}}={\frac{{{4}}}{{{15}}}}\)

a) The given surface \(\displaystyle{x}^{{2}}+{y}^{{2}}\leq{z}\leq{1}\) is the paraboloid \(\displaystyle{x}^{{2}}+{y}^{{2}}={z}\) under the plane z=1. In cylinderical co-ordinates the region bounded by the surface is given by: \(\displaystyle{E}={\left[{\left({r},\theta,{z}\right)}:{0}\leq\theta\leq{2}\pi,{0}\leq{r}\leq{1},{r}^{{2}}\leq{z}\leq{1}\right]}\)

By Gauss theorem, \(\displaystyle\int\int_{{S}}{F}\cdot{d}{s}=\int\int\int_{{E}}\div{F}{d}{V}\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{1}}}{\int_{{{r}^{{2}}}}^{{1}}}{r}{\cos{\theta}}{r}\ {\left.{d}{z}\right.}{d}{r}{d}\theta\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{1}}}{r}^{{2}}{\cos{\theta}}{\left({1}-{r}^{{2}}\right)}{d}{r}{d}\theta\)

\(\displaystyle={\int_{{0}}^{{{2}\pi}}}{\cos{\theta}}{d}\theta{\int_{{0}}^{{1}}}{\left({r}^{{2}}-{r}^{{4}}\right)}{d}{r}\)

\(\displaystyle{{\left[{\sin{\theta}}\right]}_{{0}}^{{{2}\pi}}}{{\left[{\frac{{{r}^{{3}}}}{{{3}}}}-{\frac{{{r}^{{5}}}}{{{5}}}}\right]}_{{0}}^{{t}}}={0}\)

b) Note that here we have an extra condition \(\displaystyle{x}\geq{0}\) which means that we have that portion of paraboloid under the plane which lies in front ofyz-plane. Then the region boundelby this surface is:

\(\displaystyle{E}={\left[{\left({r},\theta,{z}\right]}\right)}:-{\frac{{\pi}}{{{2}}}}\leq\theta\leq{\frac{{\pi}}{{{2}}}},{0}\leq{r}\leq{1},{r}^{{2}}\leq{z}\leq{1}{]}\)

\(\displaystyle\int\int{F}\cdot{d}{s}=\int\int\int_{{E}}\div{F}{d}{V}\)

Then

\(\displaystyle={\int_{{-{\frac{{\pi}}{{{2}}}}}}^{{\frac{{\pi}}{{{2}}}}}}{\int_{{0}}^{{1}}}{\int_{{{r}^{{2}}}}^{{1}}}{r}{\cos{\theta}}{r}{\left.{d}{z}\right.}{d}{r}{d}\theta\)

\(\displaystyle={{\left[{\sin{\theta}}\right]}_{{-{\frac{{\pi}}{{{2}}}}}}^{{{\frac{{\pi}}{{{2}}}}}}}{{\left[{\frac{{{r}^{{3}}}}{{{3}}}}-{\frac{{{r}^{{5}}}}{{{5}}}}\right]}_{{0}}^{{1}}}\)

\(\displaystyle={2}\times{\frac{{{2}}}{{{15}}}}={\frac{{{4}}}{{{15}}}}\)