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Chapter 4: Problem 9

Prove that if a surface \(z=f(x, y)\) is given in implicit form \(F(x, y, z)=0\), then the surface area (4.72) becomes $$ \iint_{R_{1 y}} \frac{\sqrt{F_{x}^{2}+F_{y}^{2}+F_{z}^{2}}}{\left|F_{z}\right|} d x d y . $$

Short Answer

Expert verified
Question: Prove that the surface area of the surface \(z = f(x, y)\) defined implicitly by the equation \(F(x, y, z) = 0\) can be calculated using the given formula: $$ \iint_{R_{1 y}} \frac{\sqrt{F_{x}^{2}+F_{y}^{2}+F_{z}^{2}}}{\left|F_{z}\right|} d x d y. $$ Answer: The surface area of the surface can be calculated by expressing the surface parametrically as \(r(u, v) = (x, y, z) = (u, v, f(u, v))\), finding the partial derivatives of the parametric representation, computing the surface area element using formula (4.72), rewriting the surface area element in terms of \(F(x, y, z)\), and finally integrating over the region. This results in the given formula $$ \iint_{R_{1 y}} \frac{\sqrt{F_x^2 + F_y^2 + F_z^2}}{|F_z|} dxdy $$ for the surface area.

Step by step solution

01

Start with the parametric representation of the surface

Let's represent the given surface parametrically as \(r(u, v) = (x, y, z) = (u, v, f(u, v))\).
02

Find partial derivatives of the parametric representation

We need to find the partial derivatives of the parametric representation with respect to \(u\) and \(v\). We can write: $$ r_u = \frac{\partial r}{\partial u} = \left(1, 0, \frac{\partial f}{\partial u}\right) $$ and $$ r_v = \frac{\partial r}{\partial v} = \left(0, 1, \frac{\partial f}{\partial v}\right) $$
03

Compute the surface area element (4.72)

The formula for the surface area of a parametric surface is given by: $$ dS = \|r_u \times r_v\| dudv $$ Compute the cross product \(r_u \times r_v\): $$ r_u \times r_v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & \frac{\partial f}{\partial u} \\ 0 & 1 & \frac{\partial f}{\partial v} \\ \end{vmatrix} = \left(-\frac{\partial f}{\partial u}, -\frac{\partial f}{\partial v}, 1\right) $$ Now calculate the magnitude of the cross product, \(\|r_u \times r_v\|\): $$ \|r_u \times r_v\| = \sqrt{\left(-\frac{\partial f}{\partial u}\right)^2 + \left(-\frac{\partial f}{\partial v}\right)^2 + 1^2} $$
04

Rewrite the surface area element in terms of \(F(x, y, z)\)

We need to express the surface area element in terms of the implicit function \(F(x, y, z)\). Note that since \(F(x, y, z) = 0\), we can use the chain rule to obtain the following expressions: $$ F_x = \frac{\partial F}{\partial x} = \frac{\partial F}{\partial z} \frac{\partial z}{\partial x} = -\frac{\partial f}{\partial u} $$ and similarly, $$ F_y = \frac{\partial F}{\partial y} = \frac{\partial F}{\partial z} \frac{\partial z}{\partial y} = -\frac{\partial f}{\partial v} $$ Replacing in the expression for the magnitude of the cross product, we obtain: $$ \|r_u \times r_v\| = \sqrt{F_x^2 + F_y^2 + 1} $$
05

Write the surface area formula

Using our results from Step 4, the surface area integral can be written as $$ \iint_{R_{1 y}} \| r_u \times r_v \| dudv = \iint_{R_{1 y}} \frac{\sqrt{F_x^2 + F_y^2 + 1}}{F_z} dxdy $$ This is the desired formula for the surface area: $$ \iint_{R_{1 y}} \frac{\sqrt{F_x^2 + F_y^2 + F_z^2}}{|F_z|} dxdy $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parametric Representation
When dealing with surfaces, the concept of parametric representation is quite useful. It allows us to express the surface in terms of two parameters, usually denoted as \(u\) and \(v\), rather than directly as \(x\), \(y\), and \(z\). In this scenario, a surface is given by the implicit function \(F(x, y, z) = 0\), and we express it through a parametric function \(r(u, v)\).

The parametric representation for the surface \(z = f(x, y)\) becomes \(r(u, v) = (x, y, f(x,y)) = (u, v, f(u, v))\). This means \(x = u\), \(y = v\), and \(z = f(u, v)\). Breaking the surface into a parametric form allows us to conduct operations like differentiation and integration more effectively.

This method also simplifies the process of calculating surface areas. Instead of directly working with \(x\), \(y\), and \(z\), we manipulate the parameters \(u\) and \(v\) to capture the essence of the surface.
Cross Product
The cross product is a vector operation that returns a vector perpendicular to two given vectors in space. It is crucial when dealing with surfaces because it assists in determining the orientation and area. To find the surface area of a parametric surface, we first take the cross product of its partial derivatives.

For our parametric surface \(r(u, v) = (u, v, f(u, v))\), we obtain the partial derivatives: \(r_u = (1, 0, f_u)\) and \(r_v = (0, 1, f_v)\). We then take their cross product:
  • \(r_u \times r_v = \left(-\frac{\partial f}{\partial u}, -\frac{\partial f}{\partial v}, 1\right)\)
The result is a normal vector to the surface, describing its orientation.

Finding the magnitude of this vector, \(||r_u \times r_v||\), gives us the surface area element from the parametric representation.
Chain Rule
The chain rule connects differentials of implicitly defined functions. It allows us to relate changes in function outputs with respect to changes in the function inputs, even when the function is complex or composed.

Here, we're given an implicit function \(F(x, y, z) = 0\) and need to express it in terms of partial derivatives. For a specific derivative like \(F_x\), the chain rule gives us this:
  • \( F_x = \frac{\partial F}{\partial x} = \frac{\partial F}{\partial z} \frac{\partial z}{\partial x} = -\frac{\partial f}{\partial u}\)
This shows how \(F_x\) relates to the partial derivative of \(f\) with respect to \(x\). Similar applications of the chain rule give us other derivatives like \(F_y\) and \(F_z\).

Mastering the chain rule aids in transforming complicated differential relationships, especially critical in geometry-related calculus problems, such as computing surface areas.
Partial Derivatives
Partial derivatives provide insight into how a multivariable function changes as one of its variables changes while keeping others constant. They are fundamental in exploring surfaces and finding tangents, normal vectors, and areas.

For the parametric surface \(z = f(u, v)\), we calculated the partial derivatives \(f_u\) and \(f_v\). These derivatives help us form the vectors \(r_u\) and \(r_v\) in finding the cross product, which ultimately assists in defining the surface area.
  • \(r_u = \left(1, 0, \frac{\partial f}{\partial u}\right)\)
  • \(r_v = \left(0, 1, \frac{\partial f}{\partial v}\right)\)
Partial derivatives also appear in the implicit context when re-expressing the surface area elements with \(F_x\), \(F_y\), and \(F_z\). They are central to applying the chain rule and understanding the nature of the surface's geometry in greater detail.

Having a good grasp of partial derivatives allows us to analyze surfaces more deeply and solve calculus problems with greater ease.

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Most popular questions from this chapter

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