Question

Set up the double integral that finds the surface area \(S\) of the given surface \(\mathcal{S},\) then use technology to approximate its value. \(\mathcal{S}\) is the paraboloid \(z=x^{2}+y^{2}\) over the triangle with vertices at (0,0),(0,1) and (1,1) .

Short answer

The surface area \(S\) is approximately 1.686 units squared.

Step-by-step solution

Parametrize the Surface

The surface is given by the equation \( z = x^2 + y^2 \). We treat \(x\) and \(y\) as parameters. So \( \mathbf{r}(x,y) = (x, y, x^2 + y^2) \) represents the parametrization of the surface.

Find the Partial Derivatives

Calculate the partial derivative of \( \mathbf{r}(x,y) \) with respect to \(x\), denoted as \( \mathbf{r}_x \), and with respect to \(y\), denoted as \( \mathbf{r}_y \):\[ \mathbf{r}_x = \left(1, 0, 2x \right) \]\[ \mathbf{r}_y = \left(0, 1, 2y \right) \]

Compute the Cross Product

Find the cross product \( \mathbf{r}_x \times \mathbf{r}_y \):\[ \mathbf{r}_x \times \mathbf{r}_y = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & 2x \ 0 & 1 & 2y \end{vmatrix} = \left(-2x, -2y, 1\right) \]

Determine the Magnitude of the Cross Product

Calculate the magnitude of the vector \( \left(-2x, -2y, 1\right) \):\[ ||\mathbf{r}_x \times \mathbf{r}_y|| = \sqrt{(-2x)^2 + (-2y)^2 + 1^2} = \sqrt{4x^2 + 4y^2 + 1} \]

Set Up the Double Integral

The surface area \(S\) is given by the double integral:\[ S = \int\int_D ||\mathbf{r}_x \times \mathbf{r}_y|| \, dA = \int_0^1 \int_0^y \sqrt{4x^2 + 4y^2 + 1} \, dx \, dy \]Here, the region \(D\) is the triangular region with vertices (0,0), (0,1), and (1,1).

Use Technology to Approximate the Integral

Use computational software or a calculator to approximate the value of the double integral. Input the integral \( \int_0^1 \int_0^y \sqrt{4x^2 + 4y^2 + 1} \, dx \, dy \) into the software to find the numerical value of \(S\).

Key concepts

Parametrization of Surfaces

Parametrization of a surface involves representing the surface as a vector function, usually with parameters like \(x\) and \(y\). This transformation is useful because it allows us to work with functions in a more flexible manner. For the paraboloid given by \(z = x^2 + y^2\), the parametrization \(\mathbf{r}(x,y) = (x, y, x^2 + y^2)\) encodes each point on the surface using just \(x\) and \(y\). By treating these variables as parameters, it becomes easier to perform calculations on the surface, such as finding its area. Parametrization turns complicated shapes into more manageable forms, facilitating further mathematical operations like integration.
In this context, understanding parametrization is key to progress with the problem. It lays the foundation for subsequent steps, such as deriving and integrating functions to determine the surface area. This transforms a 3D surface problem into a 2D parameter space problem.

Cross Product

The cross product is a mathematical operation on two vectors in three-dimensional space, producing another vector that is orthogonal to the plane containing the original vectors. When dealing with surfaces, the cross product of two partial derivative vectors yields a vector that is perpendicular to the tangent plane at a given point.
In our scenario, we compute the cross product of \( \mathbf{r}_x = (1, 0, 2x) \) and \( \mathbf{r}_y = (0, 1, 2y) \). The formula for the cross product in this context is given by:

• \( \mathbf{r}_x \times \mathbf{r}_y = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & 2x \ 0 & 1 & 2y \end{vmatrix} = (-2x, -2y, 1) \)

This resulting vector \((-2x, -2y, 1)\) is critical for determining the orientation and magnitude of the surface patch. Hence, understanding the cross product is essential for further exploration of surface properties.

Magnitude of a Vector

The magnitude of a vector is its length in space and is used to measure the 'size' of the vector independently of its direction. In the context of surface area calculation, we need the magnitude of the vector resulting from the cross product.
For our vector \((-2x, -2y, 1)\), the magnitude is calculated using the formula:
\[ ||\mathbf{r}_x \times \mathbf{r}_y|| = \sqrt{(-2x)^2 + (-2y)^2 + 1^2} = \sqrt{4x^2 + 4y^2 + 1} \]
The magnitude \(\sqrt{4x^2 + 4y^2 + 1}\) represents the rate of change in the surface area, a vital part of evaluating the integral for the surface area calculation. This magnitude plays a significant role as it becomes the integrand for our double integral, representing the surface area element.

Surface Area Calculation

Calculating the surface area of a complex shape requires setting up and evaluating a double integral. The integral leverages the parametrized form of the surface and the magnitude of the cross product calculated earlier.
The double integral is given by:

• \( S = \int\int_D ||\mathbf{r}_x \times \mathbf{r}_y|| \ dA \)

In this exercise, the region \(D\) is the triangle with vertices (0,0), (0,1), and (1,1), leading to the iterated integral:
\[ S = \int_0^1 \int_0^y \sqrt{4x^2 + 4y^2 + 1} \ dx \ dy \]
This integral allows us to sum up little patches of area across the entire surface, approximating the total area. Modern computational tools can make this calculation straightforward, providing an approximate value for the surface area \(S\). Understanding each component, from setting up the integral to using computational tools, is crucial for mastering surface area calculations of parametrized surfaces.