Question

Special case of surface integrals of scalar-valued functions Suppose 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 \(\iint_{S} f(x, y, z) d S=\iint_{R} f(x, y, g(x, y)) \sqrt{z_{x}^{2}+z_{y}^{2}+1} d A\)

Short answer

Given a surface S given by z = g(x, y), show that \(\mathbf{t}_{x} \times \mathbf{t}_{y} = \left\langle -z_x, -z_y, 1 \right\rangle\) and that \(\iint_{S} f(x, y, z) dS =\iint_{R} f(x, y, g(x, y))\sqrt{z_x^2 + z_y^2 + 1} dA\). We first find the tangent vectors of the surface S with respect to x and y, which are \(\mathbf{t}_x = \left\langle 1, 0, z_x \right\rangle\) and \(\mathbf{t}_y = \left\langle 0, 1, z_y \right\rangle\). When we take their cross product, we get \(\mathbf{t}_x \times \mathbf{t}_y = \left\langle -z_x, -z_y, 1 \right\rangle\). To prove the surface integral, we rewrite it in terms of the vector function using the formula \(\iint_{S} f(x, y, z) dS =\iint_{R} f(x, y, g(x, y)) \|\mathbf{t}_x \times \mathbf{t}_y\| dA\). Substituting the values, we get \(\iint_{S} f(x, y, z) dS =\iint_{R} f(x, y, g(x, y))\sqrt{z_x^2 + z_y^2 + 1} dA\).

Step-by-step solution

Find the tangent vectors

To find the tangent vectors of the surface \(S\), we need to take the partial derivative of the position vector with respect to \(x\) and \(y\). The position vector for the surface \(S\) can be written as \(\mathbf{r}(x,y)=\langle x, y, g(x,y)\rangle\). Now, we will take the partial derivatives. \(\mathbf{t}_x = \frac{\partial \mathbf{r}}{\partial x} = \left\langle 1, 0, \frac{\partial g}{\partial x} \right\rangle = \left\langle 1, 0, z_x \right\rangle\) \(\mathbf{t}_y = \frac{\partial \mathbf{r}}{\partial y} = \left\langle 0, 1, \frac{\partial g}{\partial y} \right\rangle = \left\langle 0, 1, z_y \right\rangle\)

Find the cross product

Now that we have the tangent vectors, we can find their cross product as follows: \(\mathbf{t}_x \times \mathbf{t}_y = \left\langle -z_x, -z_y, 1 \right\rangle\) (We can directly compute the cross product by noting that the \(x\) and \(y\) components are orthogonal, hence we just need to find the \(z\) component, which is 1)

Rewrite the surface integral

To rewrite the surface integral in terms of the vector function, we can use the following formula: \(\iint_{S} f(x, y, z) dS =\iint_{R} f(x, y, g(x, y)) \|\mathbf{t}_x \times \mathbf{t}_y\| dA\)

Substitute values and simplify

Now we can substitute the values we found in the previous steps into this formula: \(\iint_{S} f(x, y, z) dS =\iint_{R} f(x, y, g(x, y)) \|\left\langle -z_x, -z_y, 1 \right\rangle\| dA\) Recall that the magnitude of a vector \(\left\langle a, b, c \right\rangle\) is \(\sqrt{a^2 + b^2 + c^2}\). Therefore, we can simplify the expression as follows: \(\iint_{S} f(x, y, z) dS =\iint_{R} f(x, y, g(x, y))\sqrt{z_x^2 + z_y^2 + 1} dA\) This completes the proof.

Key concepts

Scalar-Valued Functions

In mathematical terms, a scalar-valued function is a regular function that outputs a single numerical value for every input. These functions map tuples of real numbers into a single real value. When analyzing surfaces, scalar-valued functions often define height or elevation based on coordinate pairs in a plane.
Imagine a mountainous terrain where each point in the landscape corresponds to a specific height. This height function, expressed as \( z = g(x, y) \), is scalar-valued because, for any pair \((x, y)\), there's a unique scalar \(z\) denoting the height.
- Surface integrals involve scalar-valued functions of three variables, typically represented as \(f(x, y, z)\).
- These functions are essential in calculating quantities distributed over a surface, such as mass or charge.
Understanding the role of scalar-valued functions helps in comprehending how surfaces interact with the amounts they may represent.

Tangent Vectors

Tangent vectors are fundamental in calculus, especially when dealing with surfaces. They represent directions along a surface and help in analyzing how the surface changes at each point. Given a surface \(z = g(x, y)\), tangent vectors are obtained by differentiating the position vector of the surface with respect to its parameters.
The position vector \( \mathbf{r}(x, y) = \langle x, y, g(x, y) \rangle \) translates the coordinates of each point on the surface into a vector form. To find tangent vectors, we calculate the partial derivatives:

• \( \mathbf{t}_x = \left\langle 1, 0, \frac{\partial g}{\partial x} \right\rangle = \left\langle 1, 0, z_x \right\rangle \) - representing the change along the x-direction.
• \( \mathbf{t}_y = \left\langle 0, 1, \frac{\partial g}{\partial y} \right\rangle = \left\langle 0, 1, z_y \right\rangle \) - representing the change along the y-direction.

These vectors help in constructing the surface's tangent plane, which approximates the surface's behavior near a given point, giving a linear approximation of the surface.

Cross Product

The cross product is a mathematical operation yielding a vector perpendicular to two given vectors. This concept is crucial in finding the orientation of planes or surfaces. For the surface \( z = g(x, y) \), the cross product of its tangent vectors gives another vector, which is perpendicular to the surface itself.
With tangent vectors \( \mathbf{t}_x = \langle 1, 0, z_x \rangle \) and \( \mathbf{t}_y = \langle 0, 1, z_y \rangle \), their cross product \( \mathbf{t}_x \times \mathbf{t}_y \) can be derived:

• Calculate \( \left\langle 1, 0, z_x \right\rangle \times \left\langle 0, 1, z_y \right\rangle = \langle -z_x, -z_y, 1 \rangle \).

This result shows a vector normal to the surface, crucial for surface integrals. Specifically, its magnitude \( \sqrt{z_x^2 + z_y^2 + 1} \) contributes directly to computing the surface integral across the region \(R\). Understanding cross products is key to visualizing how two planes interact spatially, providing insight into the geometry of surfaces.

Partial Derivatives

To understand how a surface defined by a function \( z = g(x, y) \) behaves, partial derivatives come into play. They measure how the function changes as each variable \( x \) or \( y \) changes independently, while the other is held constant.
Partial derivatives are noted as \( z_x \) and \( z_y \), representing the rate of change of \( z \) with respect to \( x \) and \( y \), respectively. They form a foundational step in deriving tangent vectors, aiding in understanding the local geometry of surfaces.
- The partial derivative \( \frac{\partial g}{\partial x} \) captures the change in \( z \) as we vary \( x \); it aligns with the x-component of the tangent vector.
- Similarly, \( \frac{\partial g}{\partial y} \) does the same with respect to \( y \), aligning with the y-component.These derivatives are instrumental in constructing the gradient, determining surface orientation, and directly inputting into the cross product for calculating normals and surface integrals. Mastering partial derivatives gives you the mathematical tools needed to dissect and analyze surfaces thoroughly.