/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 Find the area of the surface cut... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the area of the surface cut from the paraboloid \(x^{2}+y^{2}-z=\) 0 by the plane \(z=2\) .

Short Answer

Expert verified
The area is the result of evaluating \(\int_{0}^{2\pi}\int_{0}^{\sqrt{2}} r \sqrt{1 + 4r^2} \, dr \, d\theta\).

Step by step solution

01

Understanding the Problem

We need to find the surface area of a section of a 3D shape, a paraboloid, which is cut by a plane. The paraboloid is defined by the equation \(x^2 + y^2 - z = 0\), and the plane is defined by \(z = 2\).
02

Solving for z in Paraboloid Equation

The equation of the paraboloid is \(x^2 + y^2 - z = 0\). Simplify it to express \(z\) in terms of \(x\) and \(y\): \(z = x^2 + y^2\). We will substitute \(z = 2\) to understand the shape of the cross section.
03

Finding Boundary of the Section

Substitute \(z = 2\) into \(z = x^2 + y^2\) to get \(x^2 + y^2 = 2\). This represents a circle with radius \(\sqrt{2}\) in the \(xy\)-plane. We will calculate the surface area of this circular cap.
04

Surface Element Calculation

Using the formula for surface area, \(dS = \sqrt{1 + (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2} \, dx \, dy\), find the partial derivatives \(\frac{\partial z}{\partial x} = 2x\) and \(\frac{\partial z}{\partial y} = 2y\). Substitute these into the surface area element formula.
05

Integrating for Surface Area

The surface area \(S\) is given by the integral \(S = \int \int_{D} \sqrt{1 + (2x)^2 + (2y)^2} \, dx \, dy\), where \(D\) is the disk defined by \(x^2 + y^2 \leq 2\). Convert to polar coordinates \((x = r \cos \theta, y = r \sin \theta)\) to simplify integration.
06

Setting Up and Solving the Integral

In polar coordinates, \(x^2 + y^2 = r^2\), the bounds are \(0 \leq r \leq \sqrt{2}\) and \(0 \leq \theta \leq 2\pi\). The integral becomes \(S = \int_{0}^{2\pi} \int_{0}^{\sqrt{2}} r \sqrt{1 + 4r^2} \, dr \, d\theta\). Calculate this integral to find the surface area.
07

Calculating the Integral

First solve the inner integral \(\int_{0}^{\sqrt{2}} r \sqrt{1 + 4r^2} \, dr\) using substitution, then multiply by \(2\pi\) to account for the full rotation around the circle. The result will give the total surface area.

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Key Concepts

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

Paraboloid
A paraboloid is a three-dimensional surface that can be visually compared to a stretched parabola. In this context, we're examining an elliptic paraboloid, defined by the equation \(x^2 + y^2 = z\). This shape curves upwards like a bowl. When analyzing such shapes, we often want to investigate sections of it, like understanding slices or caps, which can involve planes intersecting the paraboloid at various points.
This specific paraboloid leads us to a surface when intersecting with a plane like \(z = 2\), which creates a circular slice on this curved surface. Recognizing and working with these intersections helps us visualize the geometric understanding behind mathematical surfaces. This exercise involves finding the area of such a slice or section.
Polar Coordinates
Polar coordinates provide an alternative way of describing a point in a plane, different from the traditional Cartesian coordinates. A point is represented by its distance \(r\) from the origin and an angle \(\theta\) from the positive \(x\)-axis. Converting to polar coordinates can significantly ease the computation of integrals, especially when dealing with circular or rotationally symmetric shapes.
For example, by converting from Cartesian \((x, y)\) to polar \((r, \theta)\), the equation \(x^2 + y^2 \leq 2\) becomes a simple range for \(r\), making it advantageous for integration. This coordinate system transformation is pivotal in simplifying the integration process, as seen in handling the disk section of the paraboloid.
Surface Integral
A surface integral extends the concept of line integrals to surfaces, allowing us to calculate a variety of quantities over a surface like area, flux, or mass. In our exercise, we are interested in finding the surface area of the section of the paraboloid.
The formula for a surface integral over a surface defined by a function \(z = f(x, y)\) involves the expression \(dS = \sqrt{1 + (f_x^\prime)^2 + (f_y^\prime)^2} \, dx \, dy\), where \(f_x^\prime\) and \(f_y^\prime\) are the partial derivatives. Calculating this requires understanding how the surface bends in space, and utilizing these partial derivatives helps express these curves in the integral form.
After setting up this integral in Cartesian coordinates, converting to polar coordinates typically simplifies the integration, benefiting from the radial symmetry of this problem.
Partial Derivatives
Partial derivatives are a fundamental tool in multivariable calculus. They tell us how a function changes as one of the variables changes, keeping the others constant. For the paraboloid \(z = x^2 + y^2\), the partial derivative \(\frac{\partial z}{\partial x}\) is \(2x\) and \(\frac{\partial z}{\partial y}\) is \(2y\).
These derivatives are essential when calculating the surface area element, \(dS\). They provide insight into how steeply our surface inclines in the \(x\) and \(y\) directions. By incorporating these into the integral for surface area, it accounts for the curvature of the surface over the region of interest.
Using partial derivatives in this context extends beyond simple calculations and allows for precise mathematical modeling of surfaces and forms, making them essential in engineering and physics applications.

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Most popular questions from this chapter

Integrate \(G(x, y, z)=x+y+z\) over the portion of the plane \(2 x+2 y+z=2\) that lies in the first octant.

In Exercises \(43-46,\) use a CAS to perform the following steps to evaluate the line integrals. $$ \begin{array}{l}{\text { a. Find } d s=|\mathbf{v}(t)| d t \text { for the path } \mathbf{r}(t)=g(t) \mathbf{i}+h(t) \mathbf{j}+k(t) \mathbf{k} \text { . }} \\ {\text { b. Express the integrand } f(g(t), h(t), k(t))|\mathbf{v}(t)| \text { as a function of }} \\ {\text { the parameter } t .} \\ {\text { c. Evaluate } \int_{C} f d s \text { using Equation }(2) \text { in the text. }}\end{array} $$ $$ \begin{array}{l}{f(x, y, z)=\left(1+\frac{9}{4} z^{1 / 3}\right)^{1 / 4} ; \quad \mathbf{r}(t)=(\cos 2 t) \mathbf{i}+(\sin 2 t) \mathbf{j}+} \\ {t^{5 / 2} \mathbf{k}, \quad 0 \leq t \leq 2 \pi}\end{array} $$

Find the area of the cap cut from the sphere \(x^{2}+y^{2}+z^{2}=2\) by the cone \(z=\sqrt{x^{2}+y^{2}}.\)

a. A torus of revolution (doughnut) is obtained by rotating a circle \(C\) in the \(x z\) -plane about the \(z\) -axis in space. (See the accompanying figure.) If \(C\) has radius \(r>0\) and center \((R, 0,0),\) show that a parametrization of the torus is $$\begin{aligned} \mathbf{r}(u, v)=&((R+r \cos u) \cos v) \mathbf{i} \\\ &+((R+r \cos u) \sin v) \mathbf{j}+(r \sin u) \mathbf{k} \end{aligned}$$ where \(0 \leq u \leq 2 \pi\) and \(0 \leq v \leq 2 \pi\) are the angles in the figure. b. Show that the surface area of the torus is \(A=4 \pi^{2} R r\)

In Exercises \(43-46,\) use a CAS and Green's Theorem to find the counterclockwise circulation of the field \(F\) around the simple closed curve C. Perform the following CAS steps. $$\begin{array}{l}{\text { a. Plot } C \text { in the } x y \text { -plane. }} \\\ {\text { b. Determine the integrand }(\partial N / \partial x)-(\partial M / \partial y) \text { for the tangen- }} \\ {\text { tial form of Green's Theorem. }} \\ {\text { c. Determine the (double integral) limits of integration from }} \\ {\text { your plot in part (a) and evaluate the curl integral for the }} \\ {\text { circulation. }}\end{array}$$ $$\mathbf{F}=\left(2 x^{3}-y^{3}\right) \mathbf{i}+\left(x^{3}+y^{3}\right) \mathbf{j}, \quad C : \text { The ellipse } \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$$

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