Question

Find the surface area \(S\) of the given surface \(\mathcal{S}\). (The associated integrals are computable without the assistance of technology.) \(\mathcal{S}\) is the plane \(z=x+y\) over the circular disk, centered at the origin, with radius 2 .

Short answer

The surface area is \(4\pi\sqrt{3}\).

Step-by-step solution

Parametrize the Surface

To find the surface area of the plane, first parametrize the plane. Given the equation of the plane, we can use the parameters \(x\) and \(y\), and express the surface as \(\mathbf{r}(x, y) = (x, y, x+y)\).

Define the Region of Integration

The surface is bounded over a circular disk centered at the origin with radius 2. In polar coordinates, this disk is represented by \(0 \leq r \leq 2\) and \(0 \leq \theta < 2\pi\).

Compute the Partial Derivatives

Compute the partial derivatives of the parameterization. These are \(\mathbf{r}_x = (1, 0, 1)\) and \(\mathbf{r}_y = (0, 1, 1)\).

Calculate the Cross Product

Find the cross product \(\mathbf{r}_x \times \mathbf{r}_y\). The result is \((1, -1, 1)\).

Compute the Magnitude of the Cross Product

Find the magnitude of the cross product vector. This is given by \(|\mathbf{r}_x \times \mathbf{r}_y| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}\).

Surface Area Integral Setup

Set up the surface area integral over the disk using polar coordinates. The surface area \(S\) is the integral \(\int_0^{2\pi} \int_0^2 \sqrt{3} \cdot r \, dr \, d\theta\).

Evaluate the Integral with Respect to \(r\)

Calculate the inner integral by integrating with respect to \(r\): \(\int_0^2 \sqrt{3} \cdot r \, dr = \sqrt{3} \left[\frac{1}{2} r^2\right]_0^2 = 2\sqrt{3}\).

Solve the Integral Over \(\theta\)

Now, integrate with respect to \(\theta\): \(\int_0^{2\pi} 2\sqrt{3} \, d\theta = 2\sqrt{3} \cdot (2\pi) = 4\pi\sqrt{3}\).

Express the Final Result

The surface area of the plane over the given disk is \(S = 4\pi\sqrt{3}\).

Key concepts

Cross Product in Calculus

In calculus, the cross product is a way to multiply two vectors in three-dimensional space. It gives a result that is perpendicular to both of the original vectors. This is especially useful when dealing with surfaces and shapes, as it helps us find normal vectors to those surfaces.

For the exercise at hand, you computed the cross product of two partial derivative vectors, \( \mathbf{r}_x = (1, 0, 1) \) and \( \mathbf{r}_y = (0, 1, 1) \), to eventually determine the surface area of the plane. The cross product formula is: \[ \mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1) \]

• Calculating these for our vectors gives us: \( (1, -1, 1) \).
• This result indicates a vector perpendicular to the plane surface.

Evaluating the magnitude of this cross product vector, shown as \( \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \), is crucial to finding the surface area. This magnitude acts like a scaling factor in our integration to find the correct area.

Polar Coordinates Integration

Polar coordinates are a useful way of representing a point in the plane using a distance and an angle. When integrating over a circular region, converting to polar coordinates can simplify the calculation significantly.

In the given problem, you defined the region of integration over a circular disk with radius 2, which translates in polar coordinates to:

• Radius, \( r \), from \( 0 \) to \( 2 \)
• Angle, \( \theta \), from \( 0 \) to \( 2\pi \)

This conversion helps when setting up surface integrals, as it reflects the geometric shape more intuitively. The integration is performed over the entire circular region, which is nicely structured using polar coordinates. The surface area integral in this context becomes: \[\int_0^{2\pi} \int_0^2 \sqrt{3} \cdot r \, dr \, d\theta\]This simplifies the problem because the integration boundaries are consistent with the circular shape.

Parametrization of Planes

Parametrization involves expressing a surface in terms of two parameters, often chosen based on the geometric problem at hand. For the plane equation \( z = x + y \), we can use parameters \( x \) and \( y \), to represent the surface as a vector function: \( \mathbf{r}(x, y) = (x, y, x+y) \).

The beauty of parametrization is that it allows us to transform complicated surfaces into simpler forms that we can analyze and work with mathematically. By setting parameters this way, any point on the plane corresponds to a combination \( (x, y) \).

• This is beneficial as it translates the surface into a flat plane, facilitating further calculations, such as finding derivatives or setting up integrals.
• It unfolds the surface into a characteristic formula that directly leads to calculus operations like differentiation and integration over the plane.

Understanding parametrization is vital for solving many problems in vector calculus, making complex geometries approachable through straightforward algebraic expressions.