Solución:

\(\frac{-}{y}=\int_0^{\pi}\sin\theta\ d\theta\int_0^{\pi}r2\ dr\ \)  / \(\int_0^{\frac{\pi}{2}}d\theta\ \int_0^{\pi}rdr\)
\(\frac{ }{y}\left[-\cos\ \theta\right]\frac{\frac{\pi}{2}}{0}\)

=\(\left[\frac{r3}{3}\right]\)\(\frac{\pi}{0}\left(1\right)\frac{r3}{3}\)\(=\frac{4r}{3\pi}\)

\(\left[\theta\right]\frac{\pi}{2\ }\) \(\left[\frac{r2}{2}\right]\)\(\frac{r}{0}\)                   \(\left(\frac{\pi}{2}\right)\left(\frac{\pi2}{2}\right)\)

\(\left[-120,120\right]\)

\(\left[-\frac{2\pi}{3}\right],\left[\frac{2\pi}{3}\right]\)\(\frac{ }{x}=\int_{ }^{ }\frac{ }{x}dl\ \)/\(\int_{ }^{ }dl\)

//\(\frac{ }{y}=\int_{ }^{ }\frac{ }{y}dl\)\(\int_{ }^{ }dl\)\(\int_{ }^{ }dl\)
\(dl=Rd\theta\)

\(\frac{\pi}{x\int_{\frac{3}{\pi}}^3R^2\cos\theta d\theta\ }\)  

        _________        \(=\ R\int_{\frac{2\pi}{3}}^{\frac{2\pi}{3}}\cos\theta d\theta\ \)

\(-\frac{\pi}{3\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}Rd\theta}\)

\(\frac{ }{y}\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}\sin\theta d\theta\ \ \)/\(\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{2}}Rd\theta\)\(R\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}\sin d\theta\) /\(\int_{-\frac{2\pi}{ }}^{\frac{2\pi}{3}}d\theta\) 
\(R\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}\sin\theta d\theta\) /\(+\frac{2\pi}{3}\)

\(\frac{ }{x}R\left[\sin\theta\right]\)\(\frac{\frac{2\pi}{3}}{-\frac{2\pi}{3}}\)\(=R\sqrt{3\ }\)\(\frac{2\pi}{3}\)\(=3\sqrt{3R}\) /\(4\pi\)\(=0.128\)

/\(\left[\theta\right]\frac{\frac{2\pi}{3}}{-\frac{2\pi}{3}}=\ R\sqrt{3\ }\)\(\frac{ }{y}R\left[-\cos\theta\right]\frac{\frac{2\pi}{3}}{\frac{2\pi}{3}}\)/  \(\frac{R\left[0.5\ +\ \left(-0.5\right)\right]}{\frac{4\pi}{3}}=0\)

Resultados:

\(\left[\theta\right]\frac{\frac{2\pi}{3}}{-\frac{2\pi}{3}}=\ R\sqrt{3\ }\)\(\frac{ }{y}R\left[-\cos\theta\right]\frac{\frac{2\pi}{3}}{\frac{2\pi}{3}}\)/  \(\frac{R\left[0.5\ +\ \left(-0.5\right)\right]}{\frac{4\pi}{3}}=0\)

Problema 2°