The value of the given surface integral is 3843[tex]\pi[/tex]/[tex]\sqrt{2}[/tex] and this can be determined by doing the integration.
Given :
S is the part of the cone z2 = x2 + y2 that lies between the planes z = 4 and z = 5.
First, parameterize S as given below:
[tex]\rm r(u,v) = (x(u,v) ,y(u,v),z(u,v))=(ucos\;v,usin\;v,u)[/tex]
In the above expression the value of [tex]\rm 4\leq u\leq 5[/tex] and [tex]\rm 0\leq v\leq 2\pi[/tex].
Now, evaluate the surface integral as given below:
[tex]\rm \int\int_S x^2z^2dS=\int\limits^{2\pi}_0\int\limits^5_4 {(ucosv)^2u^2} ||r_u\times r_v||\, du\;dv[/tex]
[tex]\rm \int\int_S x^2z^2dS=\sqrt{2} \int\limits^{2\pi}_0\int\limits^5_4 {(u^5cos^2v)} \, du\;dv[/tex]
Simplify the above integration in order to get the value of the surface integral.
[tex]\rm \int\int_S x^2z^2dS=\dfrac{3843\pi}{\sqrt{2} }[/tex]
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https://brainly.com/question/14502499
