Evaluate the surface integral. s xyz ds, s is the cone with parametric equations x = u cos(v), y = u sin(v), z = u, 0 ≤ u ≤ 3, 0
≤ v ≤ π 2
2 answers:
Answer:

Step-by-step explanation:
Let 
Differentiate partially with respect to u, v.




To find f, put
in xyz.

So,

Integrate with respect to u.
![\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](https://tex.z-dn.net/?f=%5Cint%20%5Cint%20f%5C%2CdS%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Cint_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%5Csin%202v%5Cleft%20%5B%20%5Cfrac%7Bu%5E5%7D%7B5%7D%20%5Cright%20%5D_%7B0%7D%5E%7B3%7D%5C%2C%5C%2Cdv%5C%2C%5C%2C%5Cleft%20%5C%7B%20%5Cbecause%20%5Cint%20u%5En%5C%2Cdu%3D%5Cfrac%7Bu%5E%7Bn%2B1%7D%7D%7Bn%2B1%7D%20%5Cright%20%5C%7D%5C%5C%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B10%7D%28243%29%5Cint_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%5Csin%202v%5C%2C%5C%2Cdv)
Integrate with respect to v.
![\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}](https://tex.z-dn.net/?f=%5Cint%20%5Cint%20f%5C%2CdS%3D%5Cfrac%7B243%5Csqrt%7B2%7D%7D%7B10%7D%5Cleft%20%5B%20%5Cfrac%7B-%5Ccos%202v%7D%7B2%7D%20%5Cright%20%5D_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%5C%2C%5C%2C%5Cleft%20%5C%7B%20%5Cbecause%20%5Cint%20%5Csin%20v%3D-%5Ccos%20v%20%5Cright%20%5C%7D%5C%5C%3D%5Cfrac%7B243%5Csqrt%7B2%7D%7D%7B20%7D%5Cleft%20%28%20-1-1%20%5Cright%20%29%5C%5C%3D%5Cfrac%7B243%5Csqrt%7B2%7D%7D%7B10%7D)
Let's capture the parameterization of the surface

by the vector function

Then the surface element is given by


So the surface integral is equivalent to
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