Question

Miscellaneous surface integrals Evaluate the following integrals using the method of your choice. Assume normal vectors point either outward or upward. \(\iint_{S} x y z d S,\) where \(S\) is that part of the plane \(z=6-y\) that lies in the cylinder \(x^{2}+y^{2}=4\)

Short answer

Question: Evaluate the surface integral \(\iint_{S} xyz dS\), where \(S\) is the part of the plane \(z = 6-y\) that lies inside the cylinder \(x^2 + y^2 = 4\). Answer: The surface integral evaluates to \(\boxed{0}\).

Step-by-step solution

Parametrize the Surface

We first parametrize the surface given by the plane, \(z = 6 - y\), and the cylinder, \(x^2 + y^2 = 4\). One way to do this is to use cylindrical coordinates \((r, \theta, z)\), where \(x = r\cos\theta, y = r\sin\theta\), and \(z = 6 - y\). Therefore, the parametrization is given by: $$\vec{r}(r, \theta) = \begin{pmatrix} r\cos\theta \\ r\sin\theta \\ 6 - r\sin\theta \end{pmatrix}$$ We have the bounds \(0 \le r \le 2\) and \(0 \le \theta \le 2\pi\) from the equation of the cylinder \(x^2 + y^2 = 4\).

Find the Differential Area \(dS\)

To find the differential area, we will compute the cross product of the partial derivatives \(\frac{\partial \vec{r}}{\partial r}\) and \(\frac{\partial \vec{r}}{\partial \theta}\). Let's first find these partial derivatives. $$\frac{\partial \vec{r}}{\partial r} = \begin{pmatrix} \cos\theta \\ \sin\theta \\ -\sin\theta \end{pmatrix} \quad \text{and} \quad \frac{\partial \vec{r}}{\partial \theta} = \begin{pmatrix} -r\sin\theta \\ r\cos\theta \\ 0 \end{pmatrix}$$ Now, we compute their cross product: $$\frac{\partial \vec{r}}{\partial r} \times \frac{\partial \vec{r}}{\partial \theta} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos\theta & \sin\theta & -\sin\theta \\ -r\sin\theta & r\cos\theta & 0 \end{vmatrix} = \begin{pmatrix} r\cos^{2}\theta \\ r\sin^{2}\theta \\ r\cos\theta \end{pmatrix}$$ Now let's find the magnitude of the cross product to obtain the differential area: $$dS = \left\lVert \frac{\partial \vec{r}}{\partial r} \times \frac{\partial \vec{r}}{\partial \theta} \right\rVert = \sqrt{(r\cos^2\theta)^2 + (r\sin^2\theta)^2 + (r\cos\theta)^2} = r$$ The differential area \(dS\) is just \(r\).

Evaluate the Triple Integral

Now we have everything we need to evaluate the surface integral: $$\iint_{S} xyz dS = \int_{0}^{2\pi} \int_{0}^{2} (r^2\cos\theta)(r\sin\theta)(6-r\sin\theta) \, r\, dr\, d\theta$$ Let's solve the integral: $$\int_{0}^{2\pi} \int_{0}^{2} (r^3\cos\theta \sin\theta)(6-r\sin\theta) \, dr\, d\theta$$ This integral can be separated into a product of two single-variable integrals: $$\left(\int_{0}^{2\pi} \cos\theta \sin\theta \, d\theta\right) \left(\int_{0}^{2} r^3 (6-r\sin\theta) \, dr\right)$$ Integrating each part separately: $$\left[\frac{1}{2}\sin^2\theta\right]_0^{2\pi} \times \left[\frac{3}{4} r^4 - \frac{1}{5} r^5\sin\theta\right]_0^2$$ Evaluating the bounds gives the final solution: $$0 \times \left[\frac{3}{4} (2^4) - \frac{1}{5} (2^5)\sin\theta\right] = \boxed{0}$$

Key concepts

Cylindrical Coordinates

Cylindrical coordinates are a coordinate system that extends the two-dimensional polar coordinate system to three dimensions. This system is highly beneficial when dealing with problems involving cylindrical symmetry. In this coordinate system, any point in space is represented as

• \( (r, \theta, z) \).

Here,

• \( r \) is the radial distance from the z-axis,
• \( \theta \) is the angle in the xy-plane from the positive x-axis,
• and \( z \) corresponds to the height along the z-axis.

In our problem, the surface given involves a part of the plane and a cylinder, making cylindrical coordinates the most suitable choice. The equations:

• \( x = r\cos\theta \)
• \( y = r\sin\theta \)
• \( z = 6 - r\sin\theta \)
are formed based on converting the plane and cylindrical equation into this system.
This helps streamline the integration process needed to solve the surface integral.

Parametrization

Parametrization is the process of representing a surface using a set of variables, usually known as parameters. For vector surfaces, this often means defining a vector function \( \vec{r}(u, v) \) which describes points on the surface in terms of parameters \( u \) and \( v \).
In our example, using the variables \( r \) and \( \theta \) from cylindrical coordinates helps to parametrize the surface with the function:\[\vec{r}(r, \theta) = \begin{pmatrix} r\cos\theta \ r\sin\theta \ 6 - r\sin\theta \end{pmatrix}\] This form makes it easy to describe the surface as it traverses over the extents of these parameters. It breaks the complex surface into manageable parts that can be calculated through simple integral calculus.
The bounds for the parameters—\(0 \leq r \leq 2\) and \(0 \leq \theta \leq 2\pi\)—describe the full area of the cylinder on this plane.

Differential Area

Differential area \( dS \) is a small element of a surface area, essential for calculating surface integrals. To determine \( dS \) for parametric surfaces, we utilize the cross product of partial derivatives of the parametrization. The derivatives are:

• \( \frac{\partial \vec{r}}{\partial r} = \begin{pmatrix} \cos\theta \ \sin\theta \ -\sin\theta \end{pmatrix} \)
• \( \frac{\partial \vec{r}}{\partial \theta} = \begin{pmatrix} -r\sin\theta \ r\cos\theta \ 0 \end{pmatrix} \)

To find the cross product, compute:\[\frac{\partial \vec{r}}{\partial r} \times \frac{\partial \vec{r}}{\partial \theta} = \begin{pmatrix} r\cos^{2}\theta \ r\sin^{2}\theta \ r\cos\theta \end{pmatrix}\]Getting the magnitude of this vector gives the differential area:\[dS = \lVert r \cos\theta \rVert = r\]This allows for a direct calculation of the surface integral by accumulating these differential pieces over the relevant region.

Cross Product

A cross product is a binary operation on two vectors in three-dimensional space, resulting in another vector perpendicular to the plane of the input vectors. This is especially useful in determining surface normals and differential areas. The cross product of two vectors \( \vec{a} \) and \( \vec{b} \) is given by:\[\vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_{1} & a_{2} & a_{3} \ b_{1} & b_{2} & b_{3} \end{vmatrix}\]In the surface integral problem, using partial derivatives of the parametrization generates vectors along the surface direction. The cross product of these vectors yields a normal vector to the surface:

• \( \begin{pmatrix} r\cos^{2}\theta \ r\sin^{2}\theta \ r\cos\theta \end{pmatrix} \)

To find the differential area, the magnitude of the resulting vector is needed. Thus, the cross product is integral in determining the direction and scaling factors for integration on the surface.