Question

Evaluate the surface integral ∫sf⋅ ds where f=⟨2x,−3z,3y⟩ and s is the part of the sphere x2 y2 z2=16 in the first octant, with outward normal orientation away from the origin

Answer 1

To evaluate the surface integral, parametrize the surface s and calculate the dot product of f and the normal vector at each point using spherical coordinates.

Step-by-step explanation:

To evaluate the surface integral ∫sf· ds, where f=⟵({2x,-3z,3y}), and s is the part of the sphere x2+y2+z2=16 in the first octant, with outward normal orientation away from the origin, we need to parametrize the surface s and calculate the dot product of f and the normal vector of s at each point.

Since the given sphere equation is symmetric with respect to the x,y, and z axes, we can consider only the part of the sphere in the first octant, where x>0, y>0, and z>0.

To parameterize the surface s, we can use spherical coordinates as follows:

x = r sin(θ) cos(φ), y = r sin(θ) sin(φ), and z = r cos(θ), where θ is the polar angle and φ is the azimuthal angle.

Answer 2

Parameterize S by the vector function

`\vec s(u,v) = \left\langle 4 \cos(u) \sin(v), 4 \sin(u) \sin(v), 4 \cos(v) \right\rangle`

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2.

Compute the outward-pointing normal vector to S :

`\vec n = (\partial\vec s)/(\partial v) * (\partial \vec s)/(\partial u) = \left\langle 16 \cos(u) \sin^2(v), 16 \sin(u) \sin^2(v), 16 \cos(v) \sin(v) \right\rangle`

The integral of the field over S is then

`\displaystyle \iint_S \vec f \cdot d\vec s = \int_0^(\frac\pi2) \int_0^(\frac\pi2) \vec f(\vec s) \cdot \vec n \, du \, dv`

`\displaystyle = \int_0^(\frac\pi2) \int_0^(\frac\pi2) \left\langle 8 \cos(u) \sin(v), -12 \cos(v), 12 \sin(u) \sin(v) \right\rangle \cdot \vec n \, du \, dv`

`\displaystyle = 128 \int_0^(\frac\pi2) \int_0^(\frac\pi2) \cos^2(u) \sin^3(v) \, du \, dv = \boxed{\frac{64\pi}3}`