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Chapter 16: Problem 5

Integrate the given function over the given surface. \begin{equation}\begin{array}{l}{\text { Portion of plane } F(x, y, z)=z, \text { over the portion of the plane }} \\ {x+y+z=4 \text { that lies above the square } 0 \leq x \leq 1,} \\ {0 \leq y \leq 1, \text { in the } x y \text { -plane }}\end{array}\end{equation}

Short Answer

Expert verified
The surface integral is 3.

Step by step solution

01

Determine the Surface Equation

To find the equation of the plane in terms of two variables, express the coordinate not present in the projection area in terms of the other two. Since the equation of the plane is given by \(x + y + z = 4\), solve for \(z\): \( z = 4 - x - y \).
02

Set Up the Surface Integral

The surface integral of a function \(F(x, y, z) = z\) over a surface \(S\) is given by \( \iint_S F(x, y, z) \, dS \). Substitute \(z = 4 - x - y\), so we need to evaluate \( \iint_R (4 - x - y) \, dA \), where \(R\) is the region on the \(xy\)-plane.
03

Define the Region of Integration

The problem describes a square region on the \(xy\)-plane given by \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\). Thus, \(R = \{(x, y): 0 \leq x \leq 1, 0 \leq y \leq 1\}\).
04

Calculate the Surface Integral

To compute the integral, evaluate the double integral over \(R\): \( \int_0^1 \int_0^1 (4 - x - y) \, dy \, dx \). Integrate with respect to \(y\) first: \( \int_0^1 \left[ 4y - xy - \frac{y^2}{2} \right]_{y=0}^{y=1} \, dx \). This simplifies to \( \int_0^1 (4 - x - \frac{1}{2}) \, dx \).
05

Complete the Integration

Now, integrate with respect to \(x\): \( \int_0^1 \left( \frac{7}{2} - x \right) \, dx \). This integral evaluates to \( \left[ \frac{7}{2}x - \frac{x^2}{2} \right]_0^1 = \frac{7}{2} - \frac{1}{2} = 3 \). Therefore, the value of the surface integral is 3.

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Key Concepts

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

Surface Integration Techniques
Surface integrals extend the concept of integration to functions over curved surfaces, rather than flat regions. When dealing with surface integrals, a crucial step is transforming the problem from three-dimensional space to two dimensions. This makes it possible to handle using two-variable calculus. Here are the steps you'll typically follow:
  • **Expressing the Surface:** Express the variable not present in the projected region in terms of the other available variables. This reduces the number of variables and simplifies the calculation.
  • **Setting Up the Integral:** Determine the function you want to integrate over the surface and set up the integral over the region after transformation.
  • **Projection to a Plane:** Project the surface onto a coordinate plane, simplifying the evaluation of the integrals, as seen in the solution where the plane equation was projected onto the xy-plane.
Breaking down the surface into such manageable sections makes it easier to compute the necessary integrals, leading to an understandable evaluation of the integral across a given surface.
Multivariable Calculus
Multivariable calculus extends the concepts of single-variable calculus to functions of several variables, enabling the exploration of more complex surfaces and curves. It plays a vital role when working with integrals that involve more than one variable.
  • **Functions of Multiple Variables:** Just like in our exercise, where the function of interest is defined over variables x, y, and z. You express these functions in terms of two variables when calculating over a surface.
  • **Partial Derivatives:** These are derivatives of a function with respect to one variable while keeping the other variables constant. They help in determining the behavior of multivariable functions over specific directions.
  • **Understanding Coordinates:** Developing an understanding of different coordinate systems like Cartesian, cylindrical, or spherical is crucial, as it aids in the simplicity of solving multivariable functions.
By grasping these fundamentals, students can deftly handle the transition between various states of integration, paving the way for more intuitive understanding and execution of complex integrals over surfaces.
Double Integrals
The concept of double integrals allows for the integration over two-dimensional regions. This involves evaluating a function over a rectangular area or another two-dimensional shape. It is a vital tool in calculating areas and volumes under surfaces. In the context of the original exercise, a double integral is applied to the projected square region on the xy-plane:
  • **Setup of Double Integral:** The bounds for each variable must be clearly defined; here, x and y range from 0 to 1. This gives the limits of integration.
  • **Order of Integration:** While the exercise follows integrating with respect to y first, it could also be performed in the reverse order, i.e., integrating with respect to x first, depending on the symmetry and simplicity it provides.
  • **Computational Steps:** Each integral is evaluated one at a time. In this example, integrating with respect to y yields an intermediate function, followed by integration with respect to x.
Mastering double integrals is indispensable for calculating multivariable functions over specified regions, whether it's for finding accumulated quantities over a defined space or for determining areas.
Plane Equation Transformation
Plane equation transformation is the process of manipulating equations to solve complex multivariable problems more easily. The transformation often involves changing the equation of a surface to simplify integration. For our exercise:
  • **Equation Manipulation:** The problem originally presents the plane as x + y + z = 4. This is transformed into z = 4 - x - y, clearly defined in terms of x and y for ease of integration.
  • **Significance in Integration:** Transformations are important because they reshape the problem into a form where the functions and limits are easier to integrate.
  • **Geometrical Insights:** By transforming the equation, one can visualize geometric properties of the surface, such as understanding how the surface interacts with the integration bounds.
Through these transformations, solving integrals over surfaces becomes more feasible, especially when approaching integration of complex surfaces in multivariable calculus.

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Most popular questions from this chapter

The tangent plane at a point \(P_{0}\left(f\left(u_{0}, v_{0}\right), g\left(u_{0}, v_{0}\right), h\left(u_{0}, v_{0}\right)\right)\) on a parametrized surface \(\mathbf{r}(u, \boldsymbol{v})=f(u, v) \mathbf{i}+g(u, v) \mathbf{j}+h(u, v) \mathbf{k}\) is the plane through \(P_{0}\) normal to the vector \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right) \times \mathbf{r}_{v}\left(u_{0}, v_{0}\right),\) the cross product of the tangent vectors \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right)\) and \(\mathbf{r}_{v}\left(u_{0}, v_{0}\right)\) at \(P_{0}\) . Find an equation for the plane tangent to the surface at \(P_{0} .\) Then find a Cartesian equation for the surface and sketch the surface and tangent plane together. Cone The cone \(\mathbf{r}(r, \theta)=(r \cos \theta) \mathbf{i}+(r \sin \theta) \mathbf{j}+r \mathbf{k}, r \geq 0\) \(0 \leq \theta \leq 2 \pi\) at the point \(P_{0}(\sqrt{2}, \sqrt{2}, 2)\) corresponding to \((r, \theta)=(2, \pi / 4)\)

Hyperboloid of two sheets Find a parametrization of the hyperboloid of two sheets \(\left(z^{2} / c^{2}\right)-\left(x^{2} / a^{2}\right)-\left(y^{2} / b^{2}\right)=1\).

Find the area of the band cut from the paraboloid \(x^{2}+y^{2}-z=\) 0 by the planes \(z=2\) and \(z=6 .\)

Use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. \(\begin{array}{l}{\text { Paraboloid } \mathbf{F}=4 x \mathbf{i}+4 y \mathbf{j}+2 \mathbf{k} \text { outward (normal away from }} \\ {\text { the } z \text { -axis through the surface cut from the bottom of the paraboloid }} \\\ {\text z=x^{2}+y^{2} \text { by the plane } z=1}\end{array}\)

Let \(R\) be a region in the \(x y\) -plane that is bounded by a piecewise smooth simple closed curve \(C\) and suppose that the moments of inertia of \(R\) about the \(x-\) and \(y\) -axes are known to be \(I_{x}\) and \(I_{y}\) .Evaluate the integral $$\oint_{C} \nabla\left(r^{4}\right) \cdot \mathbf{n} d s$$ where \(r=\sqrt{x^{2}+y^{2}},\) in terms of \(I_{x}\) and \(I_{y}.\)
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