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Consider the surface integral (S)xyzdS\int_{(S)} xyz \, dS where SS is the cone with parametric equations \newlinex=ucosvx=u\cos v, \newliney=usinvy=u\sin v, \newlinez=uz=u\newline for 0u20 \leq u \leq 2 and 0vπ20 \leq v \leq \frac{\pi}{2}.

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Q. Consider the surface integral (S)xyzdS\int_{(S)} xyz \, dS where SS is the cone with parametric equations \newlinex=ucosvx=u\cos v, \newliney=usinvy=u\sin v, \newlinez=uz=u\newline for 0u20 \leq u \leq 2 and 0vπ20 \leq v \leq \frac{\pi}{2}.
  1. Identify Parametric Equations: Identify the parametric equations and the limits for uu and vv. The cone is given by x=ucos(v)x = u \cos(v), y=usin(v)y = u \sin(v), and z=uz = u. The limits are 0u20 \leq u \leq 2 and 0vπ20 \leq v \leq \frac{\pi}{2}.
  2. Express xyzxyz in Terms: Express xyzxyz in terms of uu and vv using the parametric equations.\newlinexyz=(ucos(v))(usin(v))(u)=u3cos(v)sin(v)xyz = (u \cos(v))(u \sin(v))(u) = u^3 \cos(v) \sin(v).
  3. Calculate Differential Surface Element: Calculate the differential surface element dSdS for the cone.\newlineThe cross product of the partial derivatives of the position vector r(u,v)=(ucos(v),usin(v),u)\mathbf{r}(u, v) = (u \cos(v), u \sin(v), u) with respect to uu and vv gives the magnitude of dSdS.\newlinedrdu=(cos(v),sin(v),1)\frac{d\mathbf{r}}{du} = (\cos(v), \sin(v), 1), drdv=(usin(v),ucos(v),0)\frac{d\mathbf{r}}{dv} = (-u \sin(v), u \cos(v), 0).\newlineCross product: iamp;jamp;k cos(v)amp;sin(v)amp;1 usin(v)amp;ucos(v)amp;0=(ucos(v)+usin(v))i(ucos(v))j+(u)k.\begin{vmatrix} i & j & k \ \cos(v) & \sin(v) & 1\ -u \sin(v) & u \cos(v) & 0\end{vmatrix} = (u \cos(v) + u \sin(v))i - (u \cos(v))j + (u)k.\newlineMagnitude of dS=(ucos(v)+usin(v))2+(ucos(v))2+(u)2dS = \sqrt{(u \cos(v) + u \sin(v))^2 + (-u \cos(v))^2 + (u)^2}.

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