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Problem 262

# Let $$\mathrm{S}$$ be the surface defined by $$\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=1$$ and let the unit normal vector function have representations directed away from the origin. Compute the integral of the function $\mathrm{f}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\mathrm{a}$$over$$\mathrm{S} .$

Expert verified
The integral of the function $$f(x,y,z) = a$$ over the surface S is given by $\iint_S f(x,y,z) \, dS = \frac{4\pi}{3}a.$
See the step by step solution

## Step 1: Find a parameterization for the unit sphere.

The surface is defined by the equation $$x^2+y^2+z^2=1$$. In spherical coordinates, we can write the coordinates of the points on the surface as: $x = \sin\theta\cos\phi,$ $y = \sin\theta\sin\phi,$ $z = \cos\theta.$ Thus, a parameterization for the unit sphere is given by: $\phi(\theta, \phi) = (\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta).$ #Step 2: Compute the partial derivatives#

## Step 2: Compute $$\phi_\theta$$ and $$\phi_\phi$$.

Now we compute the partial derivatives of $$\phi$$ with respect to $$\theta$$ and $$\phi$$: $\phi_\theta = (\cos\theta\cos\phi, \cos\theta\sin\phi, -\sin\theta),$ $\phi_\phi = (-\sin\theta\sin\phi, \sin\theta\cos\phi, 0).$ #Step 3: Compute the cross product#

## Step 3: Compute the cross product of $$\phi_\theta$$ and $$\phi_\phi$$.

Now we compute the cross product of $$\phi_\theta$$ and $$\phi_\phi$$: $\phi_\theta \times \phi_\phi = \begin{vmatrix} i & j & k \\ \cos\theta\cos\phi & \cos\theta\sin\phi & -\sin\theta \\ -\sin\theta\sin\phi & \sin\theta\cos\phi & 0 \end{vmatrix} = (\sin^2\theta, \cos\theta\sin\theta\cos\phi, \cos\theta\sin\theta\sin\phi).$ #Step 4: Set up and compute the integral#

## Step 4: Compute the surface integral of $$f(x,y,z)=a$$.

We can use the surface integral formula to compute the integral of $$f(x,y,z)=a$$ over the surface of the unit sphere: $\iint_S f(x,y,z) \, dS = \iint_D a \cdot (\phi_\theta \times \phi_\phi) \, d\theta \, d\phi.$ Notice that this integral doesn't actually depend on the location on the surface S, just the geometric properties of the surface itself. Since the unit sphere has two parameters, $$\theta$$ and $$\phi$$, the domain D is given by the range of these parameters: $$0 \le \theta \le \pi$$ and $$0 \le \phi \le 2\pi$$. Substitute the values of $$f(x, y, z) = a$$ and the cross product we computed in step 3: $\iint_D a\cdot(\sin^2\theta, \cos\theta\sin\theta\cos\phi, \cos\theta\sin\theta\sin\phi) \, d\theta\, d\phi.$ Now, we will compute the iterated integral by integrating with respect to $$\theta$$ and then with respect to $$\phi$$: $\int_{0}^{2\pi} \int_{0}^{\pi} a\cdot(\sin^2\theta, \cos\theta\sin\theta\cos\phi, \cos\theta\sin\theta\sin\phi) \, d\theta\, d\phi.$ Compute the inner integral with respect to $$\theta$$: $a\int_{0}^{2\pi}\left(\left[\frac{1}{3}\sin^3\theta \right]_0^\pi + \left[\frac{1}{2}\cos^2\theta\sin^2\theta - \sin^3\theta \right]_0^\pi + \left[\frac{1}{2}\cos^2\theta\sin^2\theta\right]_0^\pi \right)\, d\phi.$ Evaluating the inner integral, we find: $a\int_{0}^{2\pi} \left(\frac{1}{3} - \frac{1}{2} + \frac{1}{6}\right)\, d\phi.$ Now, compute the outer integral with respect to $$\phi$$: $a\left[\left(\frac{2\pi}{3}\right)\right]_0^{2\pi} = a\cdot\left(\frac{4\pi}{3}\right).$ So, the integral of the function $$f(x,y,z) = a$$ over the surface S is given by: $\boxed{\iint_S f(x,y,z) \, dS = \frac{4\pi}{3}a}.$

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