Problem 18 - Advanced Engineering Mathematics

Question

Evaluate the surface integral \(\iint_{S} G(x, y, z) d S\). \(G(x, y, z)=x+y+z ; S\) the cone \(z=\sqrt{x^{2}+y^{2}}\) between \(z=1\) and \(z=4\)

Step-by-Step Solution

Answer

The value of the surface integral is \(42\pi\sqrt{2}\).

Step 1: Parametrize the Surface

The given surface is a cone described by the equation \(z = \sqrt{x^2 + y^2}\). We can parametrize the cone using cylindrical coordinates: \(x = r \cos \theta\), \(y = r \sin \theta\), \(z = r\), where \(r\) ranges from 1 to 4 (the bounds of \(z\)), and \(\theta\) ranges from 0 to \(2\pi\).

Step 2: Calculate the Position Vector

The position vector for the surface is \(\vec{r}(r, \theta) = (r \cos \theta, r \sin \theta, r)\).

Step 3: Determine the Surface Element

To find the surface element, compute the partial derivatives \(\frac{\partial \vec{r}}{\partial r}\) and \(\frac{\partial \vec{r}}{\partial \theta}\). These are \((\cos \theta, \sin \theta, 1)\) and \((-r \sin \theta, r \cos \theta, 0)\), respectively. The normal vector is given by their cross product: \((r \cos \theta, r \sin \theta, -r)\). The magnitude of this vector, which is the differential surface area element, is \(r\sqrt{1+0+1} = r\sqrt{2}\). Therefore, the surface element is \(dS = r\sqrt{2} \; dr \; d\theta\).

Step 4: Define the Integral

The function \(G(x, y, z) = x + y + z\) becomes \(r \cos \theta + r \sin \theta + r = r(\cos \theta + \sin \theta + 1)\) in the parametrized form. Substitute this into the surface integral: \[\iint_S G(x, y, z) \, dS = \int_0^{2\pi} \int_1^4 r(\cos \theta + \sin \theta + 1) \cdot r \sqrt{2} \, dr \, d\theta.\]

Step 5: Evaluate the Inner Integral

Focus on evaluating the inner integral \(\int_1^4 r^2(\cos \theta + \sin \theta + 1) \, dr\). Integration yields: \[\int_1^4 r^2 \, dr = \frac{r^3}{3}\Big|_1^4 = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21.\] For terms \(r^2 \cos \theta\) and \(r^2 \sin \theta\), they integrate separately assuming constant \(\cos \theta\) or \(\sin \theta\).

Step 6: Evaluate the Outer Integral

The outer integral \(\int_0^{2\pi} (\cos \theta + \sin \theta + 1) \, d\theta\) needs to be evaluated separately for each term. \[\int_0^{2\pi} \cos \theta \, d\theta = 0, \quad \int_0^{2\pi} \sin \theta \, d\theta = 0, \quad \int_0^{2\pi} 1 \, d\theta = 2\pi.\] Therefore, the integral becomes \(0 + 0 + 2\pi = 2\pi\).

Step 7: Combine Results

By multiplying results from Steps 5 and 6: \(\sqrt{2} \times 21 \times 2\pi\), we obtain \(42\pi\sqrt{2}\). Thus, the value of the surface integral is \(42\pi\sqrt{2}\).

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional coordinate system that extends polar coordinates with an additional height component. This system is particularly effective for surfaces with symmetry around an axis, like cones or cylinders. In cylindrical coordinates, a point in space is represented by three values:

• \(r\): the radial distance from the z-axis,
• \(\theta\): the angle in the xy-plane from the positive x-axis,
• \(z\): the height above the xy-plane.

These coordinates are interconnected with Cartesian coordinates through the equations:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)
• \(z = z\)

For our surface, the cone, we set \(z = r\) to align with its sloping geometry.

Parametrization of Surfaces

Parametrization is essential for defining complex surfaces in simpler terms that can be easily integrated. For vector calculus, working with parameters \(r\) and \(\theta\) allows us to describe each point on the cone's surface smoothly. The position vector \(\vec{r}(r, \theta)\) is a function that maps these parameters to the surface in three-dimensional space, written as: \[\vec{r}(r, \theta) = (r \cos \theta, r \sin \theta, r)\]This equation characterizes every point on the cone between specific bounds, simplifying the otherwise complex surface into a manageable format for further calculation.

Differential Surface Area

The concept of differential surface area \(dS\) is crucial when calculating integrals over a surface. It represents a tiny 'patch' of the surface area, simplified for handling through integration. To find this in vector calculus, we derive the tangent vectors by taking partial derivatives of the parametrization. For our cone, they are:

• \(\frac{\partial \vec{r}}{\partial r} = (\cos \theta, \sin \theta, 1)\)
• \(\frac{\partial \vec{r}}{\partial \theta} = (-r \sin \theta, r \cos \theta, 0)\)

The cross product of these vectors gives the normal vector to the surface: \[(r \cos \theta, r \sin \theta, -r)\]The magnitude of this normal vector (\(r\sqrt{2}\)) forms the basis of \(dS\). Thus, the surface differential element is expressed as: \[dS = r \sqrt{2} \, dr \, d\theta\]

Vector Calculus

Vector calculus is the branch of mathematics used to deal with vector fields and differential operations on them, such as calculating surface integrals. In the context of our problem, it involves integrating the function \(G(x, y, z) = x + y + z\) over a parametrized surface area. By converting into cylindrical coordinates, \(G(x, y, z)\) transforms into \(r(\cos \theta + \sin \theta + 1)\), simplifying the integral process. The surface integral is defined over the area by integrating the function times \(dS\):\[iint_S G(x, y, z) \, dS\]To solve it, you:

• Calculate the inner integral over \(r\)
• Then evaluate the outer integral over \(\theta\)

Combining these allows us to find the total contribution of \(G(x, y, z)\) over the entire surface.