Question

Suppose a > 0 is constant and consider the parameteric surface sigma given by r(phi, theta) = a sin(phi) cos(theta)i + a sin(phi) j + a cos(phi) k. 0 lessthanorequalto theta lessthanorequalto 2 pi, 0 lessthanorequalto phi lessthanorequalto pi. (a) Directly verify algebraically that r parameterizes the sphere x^2 + y^2 + z^2 = a^2, by substituting x = a sin(phi), y = a sin(phi) sin(theta), and z = a cos(phi) into the left-hand side of the equation. (b) Find r_phi, r_theta, r_phi times r_theta, and |r_phi times r_theta|. (c) Compute the surface area of the sphere doubleintegral_sigma l dS using change of variables. Find the surface area of the band sigma cut from the paraboloid z = x^2 + y^2 by the planes z = 2 and z = 6 by first finding a parameterization for the surface and then computing doubleintegral_sigma dS. Find the flux of the field F = x^2j - xzk across the surface cut by the parabolic cylinder y = x^2, -1 lessthanorequalto x lessthanorequalto 1, by the planes z = 0 and z = 2, Your normal vector should point in the direction indicated in the figure below.

Answer 1

$\Sigma$ should have parameterization

$\vec r(\varphi,\theta)=a\sin\varphi\cos\theta\,\vec\imath+a\sin\varphi\sin\theta\,\vec\jmath+a\cos\varphi\,\vec k$

if it's supposed to capture the sphere of radius $a$ centered at the origin. ($\sin\theta$ is missing from the second component)

a. You should substitute $x=a\sin\varphi\cos\theta$ (missing $\cos\theta$ this time...). Then

$x^2+y^2+z^2=(a\sin\varphi\cos\theta)^2+(a\sin\varphi\sin\theta)^2+(a\cos\varphi)^2$

$x^2+y^2+z^2=a^2\left(\sin^2\varphi\cos^2\theta+\sin^2\varphi\sin^2\theta+\cos^2\varphi\right)$

$x^2+y^2+z^2=a^2\left(\sin^2\varphi\left(\cos^2\theta+\sin^2\theta\right)+\cos^2\varphi\right)$

$x^2+y^2+z^2=a^2\left(\sin^2\varphi+\cos^2\varphi\right)$

$x^2+y^2+z^2=a^2$

as required.

b. We have

$\vec r_\varphi=a\cos\varphi\cos\theta\,\vec\imath+a\cos\varphi\sin\theta\,\vec\jmath-a\sin\varphi\,\vec k$

$\vec r_\theta=-a\sin\varphi\sin\theta\,\vec\imath+a\sin\varphi\cos\theta\,\vec\jmath$

$\vec r_\varphi\times\vec r_\theta=a^2\sin^2\varphi\cos\theta\,\vec\imath+a^2\sin^2\varphi\sin\theta\,\vec\jmath+a^2\cos\varphi\sin\varphi\,\vec k$

$\|\vec r_\varphi\times\vec r_\theta\|=a^2\sin\varphi$

c. The surface area of $\Sigma$ is

$\displaystyle\iint_\Sigma\mathrm dS=a^2\int_0^\pi\int_0^{2\pi}\sin\varphi\,\mathrm d\theta\,\mathrm d\varphi$

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

$\displaystyle\iint_\Sigma\mathrm dS=4\pi a^2$

# # #

Looks like there's an altogether different question being asked now. Parameterize $\Sigma$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k$

with $\sqrt2\le u\le\sqrt6$ and $0\le v\le2\pi$. Then

$\|\vec s_u\times\vec s_v\|=u\sqrt{1+4u^2}$

The surface area of $\Sigma$ is

$\displaystyle\iint_\Sigma\mathrm dS=\int_0^{2\pi}\int_{\sqrt2}^{\sqrt6}u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$

The integrand doesn't depend on $v$, so integration with respect to $v$ contributes a factor of $2\pi$. Substitute $w=1+4u^2$ to get $\mathrm dw=8u\,\mathrm du$. Then

$\displaystyle\iint_\Sigma\mathrm dS=\frac\pi4\int_9^{25}\sqrt w\,\mathrm dw=\frac{49\pi}3$

# # #

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 $\Sigma$ by

$\vec t(u,v)=u\,\vec\imath+u^2\,\vec\jmath+v\,\vec k$

with $-1\le u\le1$ and $0\le v\le 2$. Take the normal vector to $\Sigma$ to be

$\vec t_u\times\vec t_v=2u\,\vec\imath-\vec\jmath$

Then the flux of $\vec F$ across $\Sigma$ is

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=\int_0^2\int_{-1}^1(u^2\,\vec\jmath-uv\,\vec k)\cdot(2u\,\vec\imath-\vec\jmath)\,\mathrm du\,\mathrm dv$

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=-\int_0^2\int_{-1}^1u^2\,\mathrm du\,\mathrm dv$

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=-2\int_{-1}^1u^2\,\mathrm du=-\frac43$

If instead the direction is toward the origin, the flux would be positive.