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Chapter 16: Problem 50

Find the area of the surfaces. The surface cut from the "nose" of the paraboloid \(x=1-\) \(y^{2}-z^{2}\) by the \(y z\) -plane

Short Answer

Expert verified
The surface area is \(2\pi\).

Step by step solution

01

Understand the Surface Equation

The surface equation of the paraboloid is given by \( x = 1 - y^2 - z^2 \). This describes a paraboloid that opens along the positive x-axis. We are tasked with finding the surface area of this shape when it is cut by the \( yz \)-plane, which occurs when \( x = 0 \). Upon setting \( x = 0 \), we have the equation \( 1 - y^2 - z^2 = 0 \). Simplifying, we find \( y^2 + z^2 = 1 \). This is the equation of a circle with radius 1 centered at the origin (0,0) in the \( yz \)-plane.
02

Parameterize the Surface

To find the area of the surface defined by \( y^2 + z^2 = 1 \), we can parameterize it as a circle. Using polar coordinates, let \( y = \cos(\theta) \) and \( z = \sin(\theta) \), where \( 0 \leq \theta \leq 2\pi \). These parameterizations will help us describe the surface as a function of \( \theta \).
03

Calculate the Differential Area Element

We need the differential area element \( dA \) for calculus. In this case, the differential area element on the curve traced by the parameterization involves the derivatives of the parameterization with respect to \( \theta \). The arc length \( ds \) along the parameterized circle is expressed as \( ds = \sqrt{(-\sin(\theta))^2 + (\cos(\theta))^2} \, d\theta = d\theta \), since the derivative is the unit circle.
04

Integrate over the Circle

To find the total area of the circle (which is part of the surface), we integrate the arc length around the full circle. The integral becomes \( \int_0^{2\pi} 1 \, d\theta = [\theta]_0^{2\pi} = 2\pi \). Thus, the total surface area is \( 2\pi \), which is the circumference of the circle with radius 1.

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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 unique and fascinating three-dimensional shape. It's like a smooth bowl that extends infinitely. Imagine slicing a carrot at an angle, and you'll get an idea of its form. Specifically, a paraboloid can open along one of the coordinate axes. In our exercise, the paraboloid is defined by the equation \( x = 1 - y^2 - z^2 \). This indicates that the shape is opening along the positive x-axis.
To better understand, picture that the paraboloid's nose is facing the positive x-direction. This orientation gives it a distinct symmetry and allows interesting interplay with standard coordinate planes.
  • The structure of the paraboloid is symmetrical with respect to the x-axis.
  • The equation \( x = 1 - y^2 - z^2 \) shows that x is dependent on both y and z. Any change in y or z affects the value of x.
  • The paraboloid cuts through the coordinate planes and sends out an infinite surface, like ripples in a pond expanding outwardly.
Surface Area
Finding the surface area of the paraboloid involves a mix of geometric intuition and calculus. For this specific problem, we focus on the region where the paraboloid is sliced by the \( yz \)-plane.
The surface area we are interested in spans a circle in the \( yz \)-plane. To gain perspective:
  • The circle formed is due to setting \( x=0 \) (because it's the part cut by the \( yz \)-plane).
  • After rearranging the paraboloid equation when \( x=0 \), we end up with \( y^2 + z^2 = 1 \).
  • This equation represents a circle with radius 1 centered at the origin in the \( yz \)-plane.
Consequently, the surface area is equivalent to determining the circumference of this circle, which is a vital process in calculating areas involving curved surfaces.
Polar Coordinates
Polar coordinates simplify dealing with circular shapes. They are particularly powerful in translating curves into more manageable terms.
In the context of our exercise, we use polar coordinates to parameterize the circle formed by the paraboloid in the \( yz \)-plane:
  • We set \( y = \cos(\theta) \) and \( z = \sin(\theta) \).
  • The angle \( \theta \) sweeps the full circle from 0 to \( 2\pi \).
This parameterization enables us to transform the equation \( y^2 + z^2 = 1 \) into manageable equations for y and z using trigonometry, leveraging the relationships intrinsic to a unit circle. The integration over these parameters then helps compute the desired surface area effectively. It's like using a perfect tool suitable for the circular nature of the problem at hand.
Parameterization
Parameterization converts complex surfaces into easily manageable equations. It entails expressing one or more variables in terms of others, creating a bridge between signature shapes and functional expressions.
For the given exercise, parameterization plays a crucial role:
  • It converts the circle \( y^2 + z^2 = 1 \) into trigonometric functions \( y = \cos(\theta) \) and \( z = \sin(\theta) \), simplifying calculations.
  • The parameter \( \theta \) runs from 0 to \( 2\pi \), thereby covering the entire circle.
  • This method permits the integration of differential area elements over the complete circle, allowing us to find the surface area as \( 2\pi \), the circumference of the circle.
By parameterizing, we reduce the complexity of calculating traditional surface areas into more streamlined integration problems. It's like breaking down a large task into smaller, more manageable steps.

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Most popular questions from this chapter

Area as a line integral Show that if \(R\) is a region in the plane bounded by a piecewise smooth, simple closed curve \(C,\) then $$ R=\oint_{C} x d y=-\oint_{C} y d x $$

Green's first formula Suppose that \(f\) and \(g\) are scalar functions with continuous first- and second-order partial derivatives throughout a region \(D\) that is bounded by a closed piecewise smooth surface \(S .\) Show that $$\iint_{S} f \nabla g \cdot \mathbf{n} d \sigma=\iiint_{D}\left(f \nabla^{2} g+\nabla f \cdot \nabla g\right) d V$$ Equation \((10)\) is Green's first formula. (Hint: Apply the Divergence Theorem to the field \(\mathbf{F}=f \nabla g . )\)

Use a CAS to perform the following steps for finding the work done by force \(\mathbf{F}\) over the given path: a. Find \(d \mathbf{r}\) for the path \(\mathbf{r}(t)=g(t) \mathbf{i}+h(t) \mathbf{j}+k(t) \mathbf{k}\) b. Evaluate the force \(\mathbf{F}\) along the path. c. Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) $$ \begin{aligned} \mathbf{F} &=\frac{3}{1+x^{2}} \mathbf{i}+\frac{2}{1+y^{2}} \mathbf{j} ; \quad \mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j} \\ 0 & \leq t \leq \pi \end{aligned} $$

Bendixson's criterion The streamlines of a planar fluid flow are the smooth curves traced by the fluid's individual particles. The vectors \(\mathbf{F}=M(x, y) \mathbf{i}+N(x, y) \mathbf{j}\) of the flow's velocity field are the tangent vectors of the streamlines. Show that if the flow takes place over a simply connected region \(R\) (no holes or missing points) and that if \(M_{x}+N_{y} \neq 0\) throughout \(R\) , then none of the streamlines in \(R\) is closed. In other words, no particle of fluid ever has a closed trajectory in \(R .\) The criterion \(M_{x}+N_{y} \neq 0\) is called Bendixson's criterion for the nonexistence of closed trajectories.

A field of tangent vectors a. Find a field \(G=P(x, y) \mathbf{i}+Q(x, y)\) in the \(x y\) -plane with the property that at any point \((a, b) \neq(0,0), G\) is a unit vector tangent to the circle \(x^{2}+y^{2}=a^{2}+b^{2}\) and pointing in the clockwise direction. b. How is G related to the spin field \(F\) in Figure 16.12\(?\)
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