Question

a) Let a surface \(S: z=f(x, y)\) be defined by an implicit equation \(F(x, y, z)=0\). Show \(\mathrm{T}\) that the surface integral \(\iint H d \sigma\) over \(S\) becomes $$ \iint_{R_{c y}} \sqrt{\left(\frac{\partial F}{\partial x}\right)^{2}+\left(\frac{\partial F}{\partial y}\right)^{2}+\left(\frac{\partial F}{\partial z}\right)^{2}} \frac{H}{\left|\frac{\partial F}{\partial z}\right|} d x d y $$ provided that \(\frac{\partial F}{\partial z} \neq 0\). i b) Prove that for the surface of part (a) with \(\mathbf{n}=\nabla F /|\nabla F|\). $$ \iint_{S}(\mathbf{v} \cdot \mathbf{n}) d \sigma=\iint_{R_{s t}}(\mathbf{v} \cdot \boldsymbol{\nabla} F) \frac{1}{\left|\frac{\partial F}{\partial z}\right|} d x d y . $$ c) Prove (5.81). d) Prove that (5.81) reduces to (5.80) when \(x=u, y=v, z=f(u, v)\).

Short answer

#Part(c): Proving Relation(5.81)# We are asked to prove a given relation (5.81) based on the assumption of the surface defined by a parametrization. However, since we are not given the context or the specific relation (5.81) to prove, we cannot address this specific part in this format. Please provide the context or the specific relation (5.81) that needs to be proved for us to complete this part of the problem. #Part(d): Proving Relation(5.80)# Similarly, we are asked to prove a relation (5.80) based on some assumptions, but the context or the specific relation (5.80) is not given to us. Therefore, we cannot address this specific part in this format. Please provide the context or the specific relation (5.80) that needs to be proved for us to complete this part of the problem.

Step-by-step solution

Differentiate Implicitly

Differentiate F(x, y, z) = 0 implicitly with respect to x, y, and z to get three equations: $$ \frac{\partial F}{\partial x} + \frac{\partial F}{\partial z} \frac{\partial z}{\partial x} = 0\\ \frac{\partial F}{\partial y} + \frac{\partial F}{\partial z} \frac{\partial z}{\partial y} = 0 $$ Since ∂F/∂z ≠ 0, we can solve these equations for ∂z/∂x and ∂z/∂y: $$ \frac{\partial z}{\partial x} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}, \quad \frac{\partial z}{\partial y} = -\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}} $$

Compute the Area Element dσ

We now compute the area element dσ for the surface S. The area element can be expressed as: $$ d\sigma = \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^{2} + \left(\frac{\partial z}{\partial y}\right)^{2}} dx dy $$ Now we substitute the expressions for ∂z/∂x and ∂z/∂y obtained in Step 1: $$ d\sigma = \sqrt{1 + \left(-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}\right)^{2} + \left(-\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}}\right)^{2}} dx dy $$ Simplify the expression for dσ: $$ d\sigma = \sqrt{\frac{\left(\frac{\partial F}{\partial x}\right)^{2} + \left(\frac{\partial F}{\partial y}\right)^{2} + \left(\frac{\partial F}{\partial z}\right)^{2}}{\left(\frac{\partial F}{\partial z}\right)^{2}}} dx dy $$ Cancel terms in the square root: $$ d\sigma = \frac{\sqrt{\left(\frac{\partial F}{\partial x}\right)^{2} + \left(\frac{\partial F}{\partial y}\right)^{2} + \left(\frac{\partial F}{\partial z}\right)^{2}}}{\left|\frac{\partial F}{\partial z}\right|} dx dy $$ Now, we have the expression for the area element dσ. To find the surface integral ∬H dσ, we only need to multiply the function H with dσ and integrate over the region R in the (x, y)-plane. #Part(b): Dot Product Evaluation#

Finding the Gradient

The given surface has the normal vector n = ∇F/|∇F|. We will use this expression and find the gradient of the function F(x, y, z). The gradient is given by: $$ \boldsymbol{\nabla} F = \left\langle \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right\rangle $$

Dot Product and Simplification

Now, we compute the dot product of the vector field v and the normal n to the surface: $$ \mathbf{v} \cdot \mathbf{n} = \mathbf{v} \cdot \frac{\boldsymbol{\nabla} F}{|\boldsymbol{\nabla} F|} $$ Multiply both sides with the denominator |∇F|: $$ \mathbf{v} \cdot \boldsymbol{\nabla} F = (\mathbf{v} \cdot \mathbf{n}) |\boldsymbol{\nabla} F| $$ We have already computed dσ from Part a. Now we need to integrate the left side of the above equation over the region Rs in the (x,y)-plane and divide by |∂F/∂z|: $$ \iint_{R_{s t}}(\mathbf{v} \cdot \boldsymbol{\nabla} F) \frac{1}{\left|\frac{\partial F}{\partial z}\right|} dx dy $$ Now, we have shown that the surface integral of the dot product of the vector field v with the normal to S can be expressed as an integral over a region R in the (x,y)-plane. Part(c) and (d) in the given #exercise# are about proving certain relations (5.81) and (5.80). However, since we are not given the context of these relations, they cannot be addressed in this format. Please provide the context required to complete these parts of the problem.

Key concepts

Implicit Differentiation

Implicit differentiation is a powerful technique used in calculus when dealing with equations not explicitly solved for one variable in terms of others. When working with surfaces in three dimensions, a surface defined implicitly by an equation like \( F(x, y, z)=0 \) does not give us \( z \) in terms of \( x \) and \( y \) directly. Instead, we must differentiate each term of the equation with respect to \( x \) and \( y \) while treating \( z \) as a function of both.

For example, differentiating \( F \) with respect to \( x \) involves applying the chain rule:
\[ \frac{\partial F}{\partial x} + \frac{\partial F}{\partial z} \frac{\partial z}{\partial x} = 0 \]
This gives us a way to express \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) in terms of \( F \)'s partial derivatives, allowing further manipulations such as computing area elements for surface integrals. By doing so, we reveal the hidden relationships between the variables and can thus navigate through complex multi-variable functions and their surfaces.

Gradient Vector

The gradient vector is central to multivariate calculus and embodies the direction and rate of the steepest ascent of a function. Represented by the symbol \( abla F \), it is composed of the partial derivatives of the function with respect to its variables.

For a surface defined implicitly by \( F(x, y, z) = 0 \), the gradient vector is:
\[ \boldsymbol{abla} F = \left\langle \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right\rangle \]
This vector has a special geometric meaning: it is perpendicular, or normal, to the surface at any given point. Therefore, it's instrumental in calculating the area of a surface as well as understanding how a function changes in three-dimensional space. The gradient can be normalized by dividing it by its magnitude, resulting in a unit normal vector, which is often used when computing integrals and flux across surfaces.

Dot Product

The dot product is an operation that takes two vectors and returns a scalar quantity, which represents the product of the vectors' magnitudes and the cosine of the angle between them. It's written as:\[ \mathbf{v} \cdot \mathbf{w} = |\mathbf{v}| |\mathbf{w}| \cos(\theta) \]
The dot product has significant implications in physics and geometry, particularly when determining whether vectors are perpendicular (orthogonal) or projecting one vector onto another.

In the context of surface integrals, the dot product is used to project one vector onto the direction of another. This is especially important when computing the flux of a vector field through a surface, where we project the vector field onto the normal to the surface. If \( \mathbf{v} \) is a vector field and \( \mathbf{n} \) is the normal vector to the surface, the integral:\[ \iint_{S}(\mathbf{v} \cdot \mathbf{n}) d\sigma \]
gives us the total 'flow' of the vector field through the surface. This operation highlights the dot product's ability to measure interaction between different directions in space.

Area Element

In multivariable calculus, an area element represents a small 'piece' of area on a surface, often denoted as \( d\sigma \). It quantifies the infinitesimal surface area over which integration is performed. The area element for a surface \( S: z=f(x, y) \) in three dimensions is usually expressed as:\[ d\sigma = \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} dx dy \]
This expression accounts for the surface's tilt in the \( x \) and \( y \) directions. For an implicitly defined surface \( F(x, y, z) = 0 \), the area element based on \( F \)'s partial derivatives is:
\[ d\sigma = \frac{\sqrt{\left(\frac{\partial F}{\partial x}\right)^{2} + \left(\frac{\partial F}{\partial y}\right)^{2} + \left(\frac{\partial F}{\partial z}\right)^{2}}}{\left|\frac{\partial F}{\partial z}\right|} dx dy \]
Using it, we can integrate scalar functions or vector fields over surfaces, enabling us to calculate physical properties like mass, charge distribution, and fluid flow across a given area.