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*\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*\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*\vec s_v\|=u√(1+4u^2)`

The surface area of
`\Sigma` is

`\displaystyle\iint_\Sigma\mathrm dS=\int_0^(2\pi)\int_(\sqrt2)^(\sqrt6)u√(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*\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.