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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} .$

Short Answer

Expert verified
The integral of the function \(f(x,y,z) = a\) over the surface S is given by \[\iint_S f(x,y,z) \, dS = \frac{4\pi}{3}a.\]
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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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Most popular questions from this chapter

Chapter 9

Verify Stokes's Theorem for the vector field $\mathrm{F}^{-}(\mathrm{x}, \mathrm{y}, \mathrm{z})\( \)=(\mathrm{z}, \mathrm{x}, \mathrm{y})$ where \(\mathrm{S}\) is defined by $\mathrm{z}=4-\mathrm{x}^{2}-\mathrm{y}^{3}, \mathrm{z} \geq 0$

Chapter 9

Let \(U\) be the interior of a closed surface \(S\). Let \(f, g\) be functions. Let \(\nabla \mathrm{f}\) be the gradient of \(\mathrm{f}\) and $\nabla^{2} \mathrm{f}\( be the divergence of the gradient of \)\mathrm{f}$. Prove: a) $\iint_{\mathrm{S}} \mathrm{f}(\nabla \mathrm{g}) \cdot \mathrm{n}^{-} \mathrm{da}=\iiint_{\mathrm{U}}\left(\mathrm{f} \nabla^{2} \mathrm{~g}+\nabla \mathrm{f} \cdot \nabla \mathrm{g}\right) \mathrm{dV}$. b) $\iint_{\mathrm{S}}(\mathrm{f} \nabla \mathrm{g}-\mathrm{g} \nabla \mathrm{f}) \cdot \mathrm{n}^{-} \mathrm{da}=\iiint_{\mathrm{U}}\left(\mathrm{f} \nabla^{2} \mathrm{~g}-\mathrm{g} \nabla^{2} \mathrm{f}\right) \mathrm{dV}$.

Chapter 9

Find the integral of the function $\mathrm{f}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\mathrm{x}\( over the surface \)z=x^{2}+y\( with \)x, y$ satisfying the inequalities \(0 \leq x \leq 1\) and \(-1 \leq y \leq 1\)

Chapter 9

Let $\mathrm{F}^{\rightarrow}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left[\left\\{-\mathrm{y} /\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)\right\\},\left\\{\mathrm{x} /\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)\right\\}, 0\right]$ Evaluate \(\oint_{C} F^{-} \cdot d r^{-}\) where \(C\) is the circle \(x^{2}+y^{2}=1 .\) Also evaluate $\int_{\mathrm{S}}\left(\operatorname{curl} \mathrm{F}^{-}\right) \cdot \mathrm{n}^{\rightarrow} \mathrm{d} \mathrm{A}$ and explain the results.

Chapter 9

Let \(\mathrm{S}\) be the hemisphere given by $\mathrm{S}:\left\\{(\mathrm{x}, \mathrm{y}, \mathrm{z}) \mid \mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}\right.$ \(=1, z>0\\}\). Let \(f\) be the function \(f(x, y, z)=x^{2} y^{2} z\). Compute the integral \(\iint_{\mathrm{S}} \mathrm{f} \mathrm{d} \mathrm{A}\).

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