Question

Evaluate the following integrals using the method of your choice. Assume normal vectors point either outward or upward. \(\iint_{S} \nabla \ln |\mathbf{r}| \cdot \mathbf{n} d S,\) where \(S\) is the hemisphere \(x^{2}+y^{2}+z^{2}=a^{2}\) for \(z \geq 0,\) and where \(\mathbf{r}=\langle x, y, z\rangle\)

Short answer

Question: Evaluate the integral \(\iint_{S} \nabla \ln |\mathbf{r}| \cdot \mathbf{n} d S\), where \(S\) is the hemisphere \(x^{2}+y^{2}+z^{2}=a^{2}\) for \(z \geq 0\) and \(\mathbf{r}=\langle x,y,z \rangle\). Answer: \(\frac{8}{3} a^2 \pi\)

Step-by-step solution

Compute the gradient of \(\ln |\mathbf{r}|\)

Recall that the gradient of a scalar function is given by \(\nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \rangle\). We have \(\ln |\mathbf{r}| = \ln \sqrt{x^2 + y^2 + z^2}\). Let's compute the gradient: \(\nabla \ln |\mathbf{r}| = \langle \frac{\partial}{\partial x} \ln \sqrt{x^2 + y^2 + z^2}, \frac{\partial}{\partial y} \ln \sqrt{x^2 + y^2 + z^2}, \frac{\partial}{\partial z} \ln \sqrt{x^2 + y^2 + z^2} \rangle\) After simplification, we get: \(\nabla \ln |\mathbf{r}| = \langle \frac{x}{x^2 + y^2 + z^2}, \frac{y}{x^2 + y^2 + z^2}, \frac{z}{x^2 + y^2 + z^2} \rangle\)

Compute the normal vector\(\mathbf{n}\) for the hemisphere

The hemisphere \(S\) can be parametrized by the following function: \(G(\theta, \phi) = \langle a\sin(\phi) \cos(\theta) , a\sin(\phi) \sin(\theta), a\cos(\phi) \rangle\) Hence, \(x = a\sin(\phi) \cos(\theta)\), \(y = a\sin(\phi) \sin(\theta)\), and \(z = a\cos(\phi)\). The domain for integration will be \(\theta \in [0, 2\pi]\), and \(\phi \in [0, \frac{\pi}{2}]\). Compute the tangent vectors for the parametrization: \(\frac{\partial G}{\partial \theta} = \langle -a\sin(\phi)\sin(\theta), a\sin(\phi)\cos(\theta), 0 \rangle\) \(\frac{\partial G}{\partial \phi} = \langle a\cos(\phi)\cos(\theta), a\cos(\phi)\sin(\theta), -a\sin(\phi) \rangle\) Now we can compute the normal vector by taking cross product of these tangent vectors: \(\mathbf{n} = \frac{\partial G}{\partial \theta} \times \frac{\partial G}{\partial \phi}\) \(\mathbf{n} = \langle a^2 \sin^2(\phi) \cos(\theta), a^2 \sin^2(\phi) \sin(\theta), a^2 \sin(\phi) \cos(\phi) \rangle\)

Set up the integral and solve

Substitute \(x\), \(y\), and \(z\) from parametrization into the gradient: \(\nabla \ln |\mathbf{r}| = \langle \frac{\sin(\phi) \cos(\theta)}{a}, \frac{\sin(\phi) \sin(\theta)}{a}, \frac{\cos(\phi)}{a} \rangle\) Now, we can take the dot product of gradient and normal vector and set the integral to find the answer: \(\iint_{S} \nabla \ln |\mathbf{r}| \cdot \mathbf{n} d S = \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} (\nabla \ln |\mathbf{r}| \cdot \mathbf{n}) a^2 \sin(\phi) d \phi d \theta\) After taking the dot product and simplifying, we get: \(\iint_{S} \nabla \ln |\mathbf{r}| \cdot \mathbf{n} d S = a^2 \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} (1 + \sin^2(\phi)) \sin(\phi) d \phi d \theta\) Now, integrate with respect to \(\phi\) and \(\theta\): \(\iint_{S} \nabla \ln |\mathbf{r}| \cdot \mathbf{n} d S = a^2 [|[-\cos(\phi) + \frac{1}{3}\cos^3(\phi)]_{0}^{\frac{\pi}{2}}| \int_{0}^{2\pi} d \theta\) This simplifies to: \(\iint_{S} \nabla \ln |\mathbf{r}| \cdot \mathbf{n} d S = a^2 (\frac{4}{3}) [2\pi - 0]\) Finally, the integral evaluates to: \(\iint_{S} \nabla \ln |\mathbf{r}| \cdot \mathbf{n} d S = \frac{8}{3} a^2 \pi\)

Key concepts

Gradient

Gradients are an essential part of vector calculus and are used when working with scalar fields. You can think of a gradient as a vector that points in the direction of the greatest rate of increase of the function. For a scalar function, the gradient is a vector made of its partial derivatives:

• Partial derivative with respect to x: how the function changes as x changes.
• Partial derivative with respect to y: how the function changes as y changes.
• Partial derivative with respect to z: reflects changes in z.

Take the function \( \ln |\mathbf{r}| = \ln \sqrt{x^2 + y^2 + z^2} \,\) its gradient is \(abla \ln |\mathbf{r}| = \langle \frac{x}{x^2 + y^2 + z^2}, \frac{y}{x^2 + y^2 + z^2}, \frac{z}{x^2 + y^2 + z^2} \rangle \). This means the steepest ascent starts in the direction of this vector. The denominator, \((x^2 + y^2 + z^2)\), scales the vector elements, providing the magnitude. Understanding gradients is crucial in evaluating surface integrals as they help determine the integrand in vector calculus problems.

Normal Vector

A normal vector is perpendicular to a surface and is used to find the orientation of the surface with respect to a certain point. In three dimensions, the cross product of two vectors lying on the surface can be used to compute the normal vector. For our hemisphere surface \( x^2 + y^2 + z^2 = a^2 \,\), the parametrization provides two tangent vectors:

• \( \frac{\partial G}{\partial \theta} = \langle -a\sin(\phi)\sin(\theta), a\sin(\phi)\cos(\theta), 0 \rangle \)
• \( \frac{\partial G}{\partial \phi} = \langle a\cos(\phi)\cos(\theta), a\cos(\phi)\sin(\theta), -a\sin(\phi) \rangle \)

By taking the cross product of these vectors, we obtain the normal vector \( \mathbf{n} = \langle a^2 \sin^2(\phi) \cos(\theta), a^2 \sin^2(\phi) \sin(\theta), a^2 \sin(\phi) \cos(\phi) \rangle \). This vector is essential for surface integrals, where the orientation of the surface dramatically affects the result.

Parametrization

Parametrization is a method used to represent a surface or curve using parameters, which allows for easier computation of integrals in calculus. For a surface like the upper hemisphere expressed by \( x^2 + y^2 + z^2 = a^2 \) for \( z \geq 0 \,\), using spherical coordinates can simplify things. The parameters \( \theta \,\) and \( \phi \,\) represent the angles in the spherical coordinate system:

• \( x = a\sin(\phi) \cos(\theta) \)
• \( y = a\sin(\phi) \sin(\theta) \)
• \( z = a\cos(\phi) \)

Where \( \theta \) is in the interval \[0, 2\pi\] and \( \phi \) is in \[0, \frac{\pi}{2}\]. This conversion from cartesian to spherical makes evaluation of surface integrals easier. Parametrization is an efficient way to handle complex shapes in calculus by reducing the problem to two dimensions.

Dot Product

The dot product of two vectors \( \mathbf{A} \,\) and \( \mathbf{B} \,\) is a measure of their mutual alignment and is calculated as \( \mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z \). This operation is essential in evaluating surface integrals since it helps determine how the gradient vector interacts with the normal vector of a surface. A significant step in the exercise is computing \(abla \ln |\mathbf{r}| \cdot \mathbf{n} \,\), where \( abla \ln |\mathbf{r}| \,\) is the direction of change, and \( \mathbf{n} \,\) represents the surface's orientation.Performing the dot product gives:

• The magnitude of the projection of the gradient in the direction of the normal.
• Shows how much the gradient "goes through" the surface.

Dot product values are crucial as they form the core of the scalar field interactions in mathematical physics and engineering, providing a fundamental tool in finding the intensity of vectors in specific directions.