Question

Evaluate the following integrals using the method of your choice. Assume normal vectors point either outward or upward. \(\iint_{S}|\mathbf{r}| d S,\) where \(S\) is the cylinder \(x^{2}+y^{2}=4,\) for \(0 \leq z \leq 8\) and where \(\mathbf{r}=\langle x, y, z\rangle\)

Short answer

#tag_title#Bottom surface#tag_content#For the bottom surface, we set \(z = 0\), then parametrize it by setting \(\mathbf{r} = \langle x, y, 0 \rangle\) and using the same polar coordinates with \(x = 2\cos\theta\) and \(y = 2\sin\theta\). Thus, \(\mathbf{r} = \langle 2\cos\theta, 2\sin\theta, 0 \rangle\). The partial derivatives are unchanged from the top surface: \(\frac{\partial\mathbf{r}}{\partial\theta} = \langle-2\sin\theta, 2\cos\theta, 0 \rangle\) \(\frac{\partial\mathbf{r}}{\partial\rho} = \langle 0, 0, 0 \rangle\) The magnitude of their cross product remains: \(|\mathbf{r}_\theta \times \mathbf{r}_\rho| = 2\) We now compute the magnitude of \(\mathbf{r}\) for the bottom surface: \(|\mathbf{r}| = \sqrt{ (2\cos\theta)^2 + (2\sin\theta)^2 + 0^2 } = \sqrt{4\cos^2\theta + 4\sin^2\theta} = 2\) The integral over the bottom surface is: \(\iint_{S_\text{bottom}} |\mathbf{r}| dS = \int_{0}^{2\pi} \int_{0}^{2} 2 \cdot 2 \rho d\rho d\theta\) #tag_title#Lateral surface#tag_content#For the lateral surface of the cylinder, we parametrize it using cylindrical coordinates, \(\mathbf{r} = \langle 2\cos\theta, 2\sin\theta, z \rangle\), with \(0 \leq z \leq 8\). The partial derivatives are: \(\frac{\partial\mathbf{r}}{\partial\theta} = \langle-2\sin\theta, 2\cos\theta, 0 \rangle\) \(\frac{\partial\mathbf{r}}{\partial z} = \langle 0, 0, 1 \rangle\) The magnitude of their cross product is: \(|\mathbf{r}_\theta \times \mathbf{r}_z| = \left|\left|-2\sin\theta, 2\cos\theta, 0 \right| \right| = 2\) We also need to compute the magnitude of \(\mathbf{r}\) for the lateral surface: \(|\mathbf{r}| = \sqrt{ (2\cos\theta)^2 + (2\sin\theta)^2 + z^2 } = \sqrt{4 + z^2}\) The integral over the lateral surface is: \(\iint_{S_\text{lateral}} |\mathbf{r}| dS = \int_{0}^{2\pi} \int_{0}^{8} \sqrt{4 + z^2} \cdot 2 dz d\theta\) #tag_title#Total surface integral#tag_content#To find the total surface integral, we add the integrals of the top, bottom, and lateral surfaces: \(\iint_S |\mathbf{r}| dS = \iint_{S_\text{top}} |\mathbf{r}| dS + \iint_{S_\text{bottom}} |\mathbf{r}| dS + \iint_{S_\text{lateral}} |\mathbf{r}| dS\) Now, compute the integrals and sum them up to find the total value.

Step-by-step solution

Top surface

For the top surface, we set \(z = 8\), then parametrize it by setting \(\mathbf{r} = \langle x, y, 8 \rangle\) and using polar coordinates with \(x = 2\cos\theta\) and \(y = 2\sin\theta\). Thus, \(\mathbf{r} = \langle 2\cos\theta, 2\sin\theta, 8 \rangle\). The partial derivatives are as follows: \(\frac{\partial\mathbf{r}}{\partial\theta} = \langle-2\sin\theta, 2\cos\theta, 0 \rangle\) \(\frac{\partial\mathbf{r}}{\partial\rho} = \langle 0, 0, 0 \rangle\) The magnitude of their cross product is: \(|\mathbf{r}_\theta \times \mathbf{r}_\rho| = \left|\left|-2\sin\theta, 2\cos\theta, 0 \right| \right| = 2\) We also need to compute the magnitude of \(\mathbf{r}\): \(|\mathbf{r}| = \sqrt{ (2\cos\theta)^2 + (2\sin\theta)^2 + 8^2 } = \sqrt{4\cos^2\theta + 4\sin^2\theta + 64} = \sqrt{4 + 64} = 2\sqrt{17}\) The integral over the top surface is: \(\iint_{S_\text{top}} |\mathbf{r}| dS = \int_{0}^{2\pi} \int_{0}^{2} 2\sqrt{17} \cdot 2 \rho d\rho d\theta\)

Key concepts

Understanding Surface Integrals

Surface integrals are a fundamental concept in vector calculus, especially when you want to calculate the flux of a vector field across a surface, or the surface area of a shape in three-dimensional space. Let's look at a typical problem involving a surface integral.

In the given problem, you need to evaluate the integral of the magnitude of the position vector field \(\mathbf{r}\) over the surface \(S\) of a cylinder. Essentially, this involves summing up the lengths (magnitudes) of every tiny vector perpendicular to each point on the surface, scaled by the area of a small patch around that point.

To compute a surface integral, you'll often parameterize the surface, convert the double integral into an iterated integral, and then use either Cartesian or other coordinate systems like cylindrical or spherical coordinates to evaluate it. For the cylinder surface in the exercise, we specifically chose cylindrical coordinates to simplify the calculations since they align neatly with the symmetry of the cylinder.

Cylindrical Coordinates System

The use of cylindrical coordinates is particularly effective for problems involving symmetry around a central axis. This coordinate system is a three-dimensional extension of polar coordinates and includes a height (\(z\)) component in addition to the radial (\(\rho\)) and angular (\(\theta\)) components.

The main advantage of cylindrical coordinates comes into play when dealing with structures like cylinders, tubes, and any object where a circular cross-section is involved. By converting to cylindrical coordinates, we often simplify integration by aligning the variables with the natural symmetry of the object.

In our exercise, the cylinder's symmetry about the z-axis makes it easier to express the points on the surface in terms of \(\theta\) and \(z\), where \(\theta\) varies from 0 to \(2\pi\) and \(z\) varies from 0 to 8.

Vector Analysis in Integrals

Vector analysis is essential when solving problems involving vector fields, like in our integral evaluation exercise. When you're tasked to integrate over a surface, it's often necessary to first parameterize the surface in terms of other variables, taking the form \(\mathbf{r}(u,v)\).

After parameterization, you then compute the partial derivatives \(\frac{\partial\mathbf{r}}{\partial u}\) and \(\frac{\partial\mathbf{r}}{\partial v}\) to find the tangent vectors to the surface at each point. The magnitude of the cross product of these tangent vectors gives you the differential area element, \(dS\), which is needed to integrate over the surface. In our example, the parameterization of the top surface of the cylinder leads to a straightforward calculation of this magnitude.

The Role of Partial Derivatives

Partial derivatives play a critical role in calculating surface integrals. When parameterizing surfaces, the calculation of the differential area element, essential for surface integrals, relies on the partial derivatives of the parameterized surface vectors.

For our cylinder, the parameterization in terms of \(\theta\) allows for the calculation of the partial derivatives with respect to both \(\theta\) and \(\rho\), though in this case, the \(\rho\) derivative turns out to be unnecessary. It's the cross product of these partial derivatives that yields the differential area element, which is then used within the integral to account for variations in surface area.

The computation of |\textbf{r}| in the exercise uses these partial derivatives and results in the magnitude of the position vector, which integrates over the top surface area of the cylinder. This example demonstrates the multidimensional choreography of using partial derivatives to inform an integral over a parameterized surface in vector calculus.