I gather you're supposed to compute the integral of [tex]G(x,y,z)=z[/tex] over a surface [tex]S[/tex] that is the part of the parabolic cylinder [tex]y=z^2[/tex] with [tex]0\le x\le2[/tex] and [tex]0\le z\le\frac{\sqrt{15}}2[/tex].
We can parameterize [tex]S[/tex] by
[tex]\vec s(x,z)=x\,\vec\imath+z^2\,\vec\jmath+z\,\vec k[/tex]
with the given constraints on [tex]x[/tex] and [tex]z[/tex]. Take the normal vector to [tex]S[/tex] to be
[tex]\vec s_x\times\vec s_z=-\vec\jmath+2z\,\vec k[/tex]
so that the surface element is
[tex]\mathrm dS=\|\vec s_x\times\vec s_z\|\,\mathrm dx\,\mathrm dz=\sqrt{1+4z^2}\,\mathrm dx\,\mathrm dz[/tex]
Then in the integral, we have
[tex]\displaystyle\iint_Sz\,\mathrm dS=\int_0^2\int_0^{\sqrt{15}/2}z\sqrt{1+4z^2}\,\mathrm dz\,\mathrm dx=\boxed{\frac{21}2}[/tex]
Following are the solution to the given expression:
[tex]\to y = z^2 \\\\\to 0 \leq X \leq 4\\\\ \to 0 \leq z \leq \frac{\sqrt{15}}{2}\\\\\to G(x, y, z) = z \\\\\to y=z^2[/tex]
[tex]\to d \sigma= \sqrt{1+ y^2_{x} + y^2_{z}} \\\\[/tex]
[tex]= \sqrt{1+ 0 + (2z)^2}\\\\= \sqrt{1+ 4z^2}\\\\[/tex]
[tex]\to \int \int_{s}\ G(x,y,z)\ d \sigma = \int^{4}_{0} \int^{\frac{\sqrt{15}}{2}}_{0} z\sqrt{1+4z^2}\ dz dx\\\\[/tex]
[tex]= \int^{4}_{0} [\frac{1}{12}(1+4z^2)^{\frac{3}{2}}]^{\frac{\sqrt{15}}{2}}_{0} \ dx\\\\= \int^{4}_{0} [\frac{1}{12}(4(\frac{\sqrt{15}}{2})^2 +1)^{\frac{3}{2}} - \frac{1}{12}(4(0)^2 +1)^{\frac{3}{2}}]\ dx\\\\= \int^{4}_{0} [\frac{21}{4}]\ dx\\\\= [\frac{21}{4}] \ \int^{4}_{0} \ dx\\\\= [\frac{21}{4}] \ [x]^{4}_{0} \\\\= [\frac{21}{4}] \ [4-0] \\\\= \frac{21}{4} \times 4 \\\\= 21\\\\[/tex]
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