Calculate the surface integral over a hemisphere defined by $x^2 + y^2 + z^2 = 9$ and z ≥ 0, with the integrand $x^2z + y^2z$. Using spherical coordinates, we parameterize the surface and compute the integral for this geometric region.

Expert Answer

To evaluate the surface integral over the hemisphere S defined by $x^2 + y^2 + z^2 = 9$ with z ≥ 0, we can use the parametric representation of the surface. In this case, we can use spherical coordinates to describe the hemisphere:

\[ x = ρsin(φ)cos(θ) \]
\[ y = ρsin(φ)sin(θ) \]
\[ z = ρcos(φ) \]

Here, ρ is the radius of the sphere, which is 3 in this case, and θ and φ are the spherical coordinates.

The surface element dS in spherical coordinates is given by:

\[ dS = ρ^2sin(φ)dφdθ \]

Now, let’s rewrite the integrand in terms of spherical coordinates:

\[ x^2z + y^2z = (ρsin(φ)cos(θ))^2(ρcos(φ)) + (ρsin(φ)sin(θ))^2(ρcos(φ)) \]

\[ = ρ^3sin(φ)cos^2(θ)cos(φ) + ρ^3sin(φ)sin^2(θ)cos(φ) \]

Now, we can set up the integral using these expressions:

\[ \iint S (x^2z + y^2z) dS \]

\[= \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} [ρ^3sin(φ)cos^2(θ)cos(φ) + ρ^3sin(φ)sin^2(θ)cos(φ)] ρ^2sin(φ) dφ dθ \]

\[ = \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} ρ^5sin(φ)cos(φ)[cos^2(θ) + sin^2(θ)] dφ dθ \]

\[ = ρ^5 \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} sin(φ)cos(φ) dφ dθ \]

Now, you can integrate with respect to φ and θ:

\[ \int_{0}^{2\pi} dθ = 2π \]

\[ \int_{0}^{\frac{\pi}{2}} sin(φ)cos(φ) dφ \]

\[ = \frac{1}{2} \left[ sin^2(φ) \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \]

So, the surface integral becomes:

\[ \iint S (x^2z + y^2z) dS \]

\[ = ρ^5 * 2π * 1/2 = 3^5 * 2π * 1/2 = 243π \]

Therefore, the value of the surface integral is 243π.

Example

Evaluate the surface integral. $\iint S (x^2z + y^2z) dS$. S is the hemisphere $x^2 +y^2 + z^2 = 4, z ≥ 0$.

Solution

\[ ρ = 2 \]
\[ 0 ≤ φ ≤ π/2 \]
\[ 0 ≤ θ ≤ 2π \]

Where ρ is the radius of the hemisphere, φ is the polar angle (ranging from 0 to π/2), and θ is the azimuthal angle (ranging from 0 to 2π).

Now, express x, y, and z in terms of spherical coordinates:

\[ x = ρsin(φ)cos(θ) \]
\[ y = ρsin(φ)sin(θ) \]
\[ z = ρcos(φ) \]

The surface element dS in spherical coordinates is given by:

\[ dS = ρ^2sin(φ)dφdθ \]

Now, rewrite the integrand (x^2z + y^2z) in terms of spherical coordinates:

\[ x^2z + y^2z = (ρsin(φ)cos(θ))^2(ρcos(φ)) + (ρsin(φ)sin(θ))^2(ρcos(φ)) \]

\[ = ρ^3sin(φ)cos^2(θ)cos(φ) + ρ^3sin(φ)sin^2(θ)cos(φ) \]

Now, set up the integral using these expressions:

\[ \iint S (x^2z + y^2z) dS \]

\[ = \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} [ρ^3sin(φ)cos^2(θ)cos(φ) + ρ^3sin(φ)sin^2(θ)cos(φ)] ρ^2sin(φ) dφ dθ \]

\[ = \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} ρ^5sin(φ)cos(φ)[cos^2(θ) + sin^2(θ)] dφ dθ \]

\[ = ρ^5 \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} ρ^5sin(φ)cos(φ) dφ dθ \]

Now, integrate with respect to φ and θ:

\[ \int_{0}^{2\pi} dθ = 2π \]

\[ \int_{0}^{\frac{\pi}{2}} sin(φ)cos(φ) dφ \]

\[ = \frac{1}{2} \left[ sin^2(φ) \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \]

So, the surface integral becomes:

\[ \iint S (x^2z + y^2z) dS \]

\[ = ρ^5 * 2π * 1/2 = 2^5 * π * 1/2 = 16π \]