Question

Suppose that a surface \(S\) is defined as \(z=g(x, y)\) on a region \(R\). Show that \(\mathbf{t}_{x} \times \mathbf{t}_{y}=\left\langle-z_{x},-z_{y}, 1\right\rangle\) and that \(\iint_{S} f(x, y, z) d S=\iint_{R} f(x, y, z) \sqrt{z_{x}^{2}+z_{y}^{2}+1} d A\).

Short answer

Question: Show that for a surface S defined as \(z = g(x, y)\) on a region R, the cross product of the tangent vectors \(\mathbf{t_x}\) and \(\mathbf{t_y}\) is equal to \(\langle -z_x, -z_y, 1 \rangle\) and the surface integral \(\iint_{S} f(x, y, z) dS\) can be expressed as \(\iint_{R} f(x, y, z) \sqrt{z_x^2 + z_y^2 + 1} dA\). Answer: We first found the tangent vectors \(\mathbf{t_x} = \langle 1, 0, z_x \rangle\) and \(\mathbf{t_y} = \langle 0, 1, z_y \rangle\) for the surface S parameterized as \(\mathbf{r}(x, y) = \langle x,y,g(x,y)\rangle\). Then we calculated the cross product \(\mathbf{t_x} \times \mathbf{t_y} = \langle -z_x, -z_y, 1 \rangle\). Finally, we used this result to express the surface integral \(\iint_{S} f(x, y, z) dS = \iint_{R} f(x, y, z) \sqrt{z_x^2 + z_y^2 + 1} dA\).

Step-by-step solution

Finding the tangent vectors \(\mathbf{t_x}\) and \(\mathbf{t_y}\) for the surface S

Given the surface S with \(z = g(x, y)\), we can parameterize it as a vector function \(\mathbf{r}(x, y) = \langle x, y, g(x, y) \rangle\). To find the tangent vectors \(\mathbf{t_x}\) and \(\mathbf{t_y}\), we need to take the partial derivatives of this vector function with respect to x and y, respectively: $$ \mathbf{t_x} = \frac{\partial \mathbf{r}}{\partial x} = \left\langle 1, 0, \frac{\partial g}{\partial x} \right\rangle = \langle 1, 0, z_x \rangle \\ \mathbf{t_y} = \frac{\partial \mathbf{r}}{\partial y} = \left\langle 0, 1, \frac{\partial g}{\partial y} \right\rangle = \langle 0, 1, z_y \rangle $$

Calculating the cross product of \(\mathbf{t_x}\) and \(\mathbf{t_y}\)

To compute the cross product \(\mathbf{t_x} \times \mathbf{t_y}\), we can use the determinant method: $$ \mathbf{t_x} \times \mathbf{t_y} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & z_x \\ 0 & 1 & z_y \end{array} \right| $$ Expanding the determinant, we get: $$ \mathbf{t_x} \times \mathbf{t_y} = \langle (0)(z_y) - (1)(z_x), (1)(z_x) - (0)(1), (1)(1) - (0)(z_y) \rangle = \langle -z_x, -z_y, 1 \rangle $$

Finding the expression for the surface integral over S

To evaluate the surface integral \(\iint_S f(x, y, z) dS\), we can use the formula: $$ \iint_S f(x, y, z) dS = \iint_R f(x, y, g(x, y)) \lVert \mathbf{t_x} \times \mathbf{t_y} \rVert dA $$ Since we already found the cross product \(\mathbf{t_x} \times \mathbf{t_y}\), we can calculate its magnitude: $$ \lVert \mathbf{t_x} \times \mathbf{t_y} \rVert = \lVert \langle -z_x, -z_y, 1 \rangle \rVert = \sqrt{(-z_x)^2 + (-z_y)^2 + 1^2} = \sqrt{z_x^2 + z_y^2 + 1} $$ Plug this result back into the surface integral formula: $$ \iint_S f(x, y, z) dS = \iint_R f(x, y, g(x, y)) \sqrt{z_x^2 + z_y^2 + 1} dA $$ This is the expression that we wanted to show. Therefore, the proof is complete.

Key concepts

Tangent Vectors

In the realm of calculus, a **tangent vector** helps us understand the direction in which a surface changes at a particular point. Imagine you have a surface defined by the equation \(z = g(x, y)\), essentially giving us the height of the surface at any point \((x, y)\).
To find the tangent vectors, which are crucial for studying the surface's behavior, we need to create a vector function parameterizing the surface: \(\mathbf{r}(x, y) = \langle x, y, g(x, y) \rangle \).

• The tangent vector in the direction of the \(x\)-axis, \(\mathbf{t_x}\), is found by taking the partial derivative of \(\mathbf{r}\) with respect to \(x\). This results in \( \langle 1, 0, z_x \rangle \), where \(z_x = \frac{\partial g}{\partial x}\).
• Similarly, the tangent vector \(\mathbf{t_y}\) in the direction of the \(y\)-axis is \( \langle 0, 1, z_y \rangle \), with \(z_y = \frac{\partial g}{\partial y}\).

Through these tangent vectors, we capture the direction and rate of change of the surface at any given point, paving the way for deeper analysis.

Cross Product

The **cross product** is a mathematical operation that allows us to compute a vector perpendicular to two given vectors. This concept is critical in surface integrals because it helps us find the normal vector to a surface.
Given two tangent vectors \(\mathbf{t_x} = \langle 1, 0, z_x \rangle\) and \(\mathbf{t_y} = \langle 0, 1, z_y \rangle\), we find their cross product using the determinant method:

• Imagine a 3x3 matrix with unit vectors \(\hat{i}, \hat{j}, \hat{k}\) in the first row, \(1, 0, z_x\) in the second row, and \(0, 1, z_y\) in the third row.
• The cross product \(\mathbf{t_x} \times \mathbf{t_y}\) is calculated by evaluating the determinant of this matrix, resulting in \( \langle -z_x, -z_y, 1 \rangle \).

This vector \( \langle -z_x, -z_y, 1 \rangle \) is orthogonal to the tangent vectors, meaning it represents the direction directly out from the surface at that point, which is essential when considering how to integrate over the surface.

Partial Derivatives

Understanding **partial derivatives** is fundamental to grasping how surfaces behave in multivariable calculus. When a surface is defined by \(z = g(x, y)\), the partial derivative \(\frac{\partial g}{\partial x}\) or \(z_x\) measures how \(z\) changes as we tweak \(x\) while keeping \(y\) fixed.
Similarly, \(\frac{\partial g}{\partial y}\) or \(z_y\) does the same but swaps the role of \(x\) and \(y\).

• These derivatives are the building blocks of tangent vectors, as they help us calculate the slope of the surface in a given direction.
• Additionally, when forming the normal vector through the cross product, these derivatives play a role since the normal vector needs information about how the surface is inclined in both directions.

Thus, partial derivatives give us a detailed, directional snapshot of how the function defining our surface behaves, which is crucial for analyzing and integrating over it.