Respuesta :
[tex]\Sigma[/tex] should have parameterization
[tex]\vec r(\varphi,\theta)=a\sin\varphi\cos\theta\,\vec\imath+a\sin\varphi\sin\theta\,\vec\jmath+a\cos\varphi\,\vec k[/tex]
if it's supposed to capture the sphere of radius [tex]a[/tex] centered at the origin. ([tex]\sin\theta[/tex] is missing from the second component)
a. You should substitute [tex]x=a\sin\varphi\cos\theta[/tex] (missing [tex]\cos\theta[/tex] this time...). Then
[tex]x^2+y^2+z^2=(a\sin\varphi\cos\theta)^2+(a\sin\varphi\sin\theta)^2+(a\cos\varphi)^2[/tex]
[tex]x^2+y^2+z^2=a^2\left(\sin^2\varphi\cos^2\theta+\sin^2\varphi\sin^2\theta+\cos^2\varphi\right)[/tex]
[tex]x^2+y^2+z^2=a^2\left(\sin^2\varphi\left(\cos^2\theta+\sin^2\theta\right)+\cos^2\varphi\right)[/tex]
[tex]x^2+y^2+z^2=a^2\left(\sin^2\varphi+\cos^2\varphi\right)[/tex]
[tex]x^2+y^2+z^2=a^2[/tex]
as required.
b. We have
[tex]\vec r_\varphi=a\cos\varphi\cos\theta\,\vec\imath+a\cos\varphi\sin\theta\,\vec\jmath-a\sin\varphi\,\vec k[/tex]
[tex]\vec r_\theta=-a\sin\varphi\sin\theta\,\vec\imath+a\sin\varphi\cos\theta\,\vec\jmath[/tex]
[tex]\vec r_\varphi\times\vec r_\theta=a^2\sin^2\varphi\cos\theta\,\vec\imath+a^2\sin^2\varphi\sin\theta\,\vec\jmath+a^2\cos\varphi\sin\varphi\,\vec k[/tex]
[tex]\|\vec r_\varphi\times\vec r_\theta\|=a^2\sin\varphi[/tex]
c. The surface area of [tex]\Sigma[/tex] is
[tex]\displaystyle\iint_\Sigma\mathrm dS=a^2\int_0^\pi\int_0^{2\pi}\sin\varphi\,\mathrm d\theta\,\mathrm d\varphi[/tex]
You don't need a substitution to compute this. The integration limits are constant, so you can separate the variables to get two integrals. You'd end up with
[tex]\displaystyle\iint_\Sigma\mathrm dS=4\pi a^2[/tex]
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Looks like there's an altogether different question being asked now. Parameterize [tex]\Sigma[/tex] by
[tex]\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k[/tex]
with [tex]\sqrt2\le u\le\sqrt6[/tex] and [tex]0\le v\le2\pi[/tex]. Then
[tex]\|\vec s_u\times\vec s_v\|=u\sqrt{1+4u^2}[/tex]
The surface area of [tex]\Sigma[/tex] is
[tex]\displaystyle\iint_\Sigma\mathrm dS=\int_0^{2\pi}\int_{\sqrt2}^{\sqrt6}u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv[/tex]
The integrand doesn't depend on [tex]v[/tex], so integration with respect to [tex]v[/tex] contributes a factor of [tex]2\pi[/tex]. Substitute [tex]w=1+4u^2[/tex] to get [tex]\mathrm dw=8u\,\mathrm du[/tex]. Then
[tex]\displaystyle\iint_\Sigma\mathrm dS=\frac\pi4\int_9^{25}\sqrt w\,\mathrm dw=\frac{49\pi}3[/tex]
# # #
Looks like yet another different question. No figure was included in your post, so I'll assume the normal vector points outward from the surface, away from the origin.
Parameterize [tex]\Sigma[/tex] by
[tex]\vec t(u,v)=u\,\vec\imath+u^2\,\vec\jmath+v\,\vec k[/tex]
with [tex]-1\le u\le1[/tex] and [tex]0\le v\le 2[/tex]. Take the normal vector to [tex]\Sigma[/tex] to be
[tex]\vec t_u\times\vec t_v=2u\,\vec\imath-\vec\jmath[/tex]
Then the flux of [tex]\vec F[/tex] across [tex]\Sigma[/tex] is
[tex]\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=\int_0^2\int_{-1}^1(u^2\,\vec\jmath-uv\,\vec k)\cdot(2u\,\vec\imath-\vec\jmath)\,\mathrm du\,\mathrm dv[/tex]
[tex]\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=-\int_0^2\int_{-1}^1u^2\,\mathrm du\,\mathrm dv[/tex]
[tex]\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=-2\int_{-1}^1u^2\,\mathrm du=-\frac43[/tex]
If instead the direction is toward the origin, the flux would be positive.