Respuesta :
Let's capture the parameterization of the surface [tex]\mathcal S[/tex] by the vector function
[tex]\mathbf s(u,v)=\langle u\cos v,u\sin v,u\rangle[/tex]
Then the surface element is given by
[tex]\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv[/tex]
[tex]\implies\mathrm dS=\sqrt 2u\,\mathrm du\,\mathrm dv[/tex]
So the surface integral is equivalent to
[tex]\displaystyle\iint_{\mathcal S}xyz\,\mathrm dS=\sqrt2\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=3}u^4\cos v\sin v\,\mathrm du\,\mathrm dv=\frac{243}{5\sqrt2}[/tex]
[tex]\mathbf s(u,v)=\langle u\cos v,u\sin v,u\rangle[/tex]
Then the surface element is given by
[tex]\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv[/tex]
[tex]\implies\mathrm dS=\sqrt 2u\,\mathrm du\,\mathrm dv[/tex]
So the surface integral is equivalent to
[tex]\displaystyle\iint_{\mathcal S}xyz\,\mathrm dS=\sqrt2\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=3}u^4\cos v\sin v\,\mathrm du\,\mathrm dv=\frac{243}{5\sqrt2}[/tex]
Answer:
[tex]\frac{243\sqrt{2}}{10}[/tex]
Step-by-step explanation:
Let [tex]r(u,v)=\left \langle u\cos v,u\sin v,u \right \rangle[/tex]
Differentiate partially with respect to u, v.
[tex]r_u=\left \langle \cos v,\sin v,1 \right \rangle\\r_v=\left \langle -u\sin v,u\cos v,0 \right \rangle[/tex]
[tex]r_u\times r_v=\left | \begin{matrix} i&j&k\\ \cos v&\sin v&1\\-u\sin v&u\cos v&0 \end{matrix} \right |=\left \langle -u\cos v,-u\sin v,u \right \rangle\\\left \|r_u\times r_v \right \|=\sqrt{u^2\cos ^2v+u^2\sin ^2v+u^2}\\=\sqrt{u^2\left (\cos ^2v+\sin ^2v \right )+u^2}\\=\sqrt{u^2+u^2}\\=\sqrt{2u^2}=\sqrt{2}u[/tex]
[tex]dS=\left \| r_u\times r_v \right \|\,du\,\,dv[/tex]
[tex]\int \int f\,dS=\int_v \int_u f\,\left \| r_u\times r_v \right \|\,du\,\,dv[/tex]
To find f, put [tex]x = u \cos(v)\,,\, y = u \sin(v)\,,\, z = u[/tex] in xyz.
[tex]f=xyz\\=(u \cos(v))( u \sin(v))(u)\\=u^3\sin v\,\cos v\\=u^3\left ( \frac{2\sin v\,\cos v}{2} \right )\\=\frac{u^3}{2}\sin 2v\,\,\left \{ \because 2\sin v\,\cos v=\sin 2v \right \}[/tex]
So,
[tex]\int \int f\,dS=\int_{0}^{\frac{\pi}{2}} \int_{0}^{3} \frac{u^3}{2}\sin 2v\,\left ( \sqrt{2}u \right )\,du\,\,dv\\=\frac{\sqrt{2}}{2}\int_{0}^{\frac{\pi}{2}} \int_{0}^{3}u^4\sin 2v\,\,du\,\,dv[/tex]
Integrate with respect to u.
[tex]\int \int f\,dS=\frac{\sqrt{2}}{2}\int_{0}^{\frac{\pi}{2}}\sin 2v\left [ \frac{u^5}{5} \right ]_{0}^{3}\,\,dv\,\,\left \{ \because \int u^n\,du=\frac{u^{n+1}}{n+1} \right \}\\=\frac{\sqrt{2}}{10}(243)\int_{0}^{\frac{\pi}{2}}\sin 2v\,\,dv[/tex]
Integrate with respect to v.
[tex]\int \int f\,dS=\frac{243\sqrt{2}}{10}\left [ \frac{-\cos 2v}{2} \right ]_{0}^{\frac{\pi}{2}}\,\,\left \{ \because \int \sin v=-\cos v \right \}\\=\frac{243\sqrt{2}}{20}\left ( -1-1 \right )\\=\frac{243\sqrt{2}}{10}[/tex]
