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Chapter 15: Problem 37

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 of the surface is \(2\pi\).

Step by step solution

01

Identify the Surface and Plane

First, we identify the three-dimensional surface and the plane that will intersect it. The paraboloid is described by the equation \(x^2 + y^2 - z = 0\), which simplifies to \(z = x^2 + y^2\). The plane is given by \(z = 2\).
02

Set Up the Intersection Equation

To find the curve of intersection between the paraboloid and the plane, set the two equations for \(z\) equal: \(x^2 + y^2 = 2\). This represents a circle in the \(xy\)-plane.
03

Parameterize the Circle

Since the intersection is a circle with equation \(x^2 + y^2 = 2\), we can parameterize it using trigonometric functions: \(x = \sqrt{2}\cos(\theta)\) and \(y = \sqrt{2}\sin(\theta)\), with \(\theta\) ranging from \(0\) to \(2\pi\).
04

Use Polar Coordinates

To find the area of the circle \(x^2 + y^2 = 2\), notice that this is a circle of radius \(\sqrt{2}\). The area \(A\) of a circle is given by the formula \(A = \pi r^2\), where \(r\) is the radius.
05

Calculate the Area

Substitute the radius \(r = \sqrt{2}\) into the area formula: \(A = \pi (\sqrt{2})^2 = 2\pi\). Therefore, the area of the surface cut from the paraboloid by the plane is \(2\pi\).

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Key Concepts

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

Understanding a Paraboloid
A paraboloid is a three-dimensional surface that looks like a stretched-out dome or a bowl. You can imagine it like a satellite dish. It's defined mathematically by equations like \(z = x^2 + y^2\). This is an upward-opening paraboloid, similar to the shape of a glass or a cup.
  • The equation \(z = x^2 + y^2\) implies that for every unit increase away from the center (where \(x = 0\) and \(y = 0\)), the value of \(z\) becomes larger. It extends infinitely upwards.
  • In the context of our problem, the paraboloid surface intersects with a horizontal plane. This intersection helps to visualize where they cut across each other.
This type of surface is common when dealing with reflective properties in physics or when designing certain architectural elements. Paraboloids automatically focus parallel incoming rays to a single point, a property used in many optical devices, along with satellite dishes.
Exploring Plane Intersection
When we talk about the intersection between a plane and a paraboloid, we are interested in finding where they meet. This intersection forms a curve or shape on the paraboloid's surface. In our problem, it's a matter of finding where the plane defined by \(z = 2\) slices through our paraboloid.
  • The plane is defined by a constant \(z\) value, which gives us a flat surface across various values of \(x\) and \(y\).
  • Setting the plane's equation equal to the paraboloid's, we get \(x^2 + y^2 = 2\), which describes a circle in the \(xy\)-plane at height \(z = 2\).
Imagine slicing a straight line through a rounded bowl; where the knife goes through will create a circular edge. That edge or boundary is the curve of intersection, and understanding this helps us to calculate areas, volumes, and other physical properties.
Parameterizing with Parametric Equations
Parametric equations are a powerful tool in calculus, used to express a set of equations through one or more variables known as parameters. In our problem, we use them to describe the circle of intersection between the paraboloid and the plane.
  • The circle \(x^2 + y^2 = 2\) can be converted using parametric equations as \(x = \sqrt{2}\cos(\theta)\) and \(y = \sqrt{2}\sin(\theta)\), where \(\theta\) varies from \(0\) to \(2\pi\).
  • Using trigonometric functions like cosine and sine ensures the parameterization describes a full circle, covering every point around it as \(\theta\) changes.
Parametric equations simplify complex calculations, allowing for easy manipulation and visualization of shapes. They provide a convenient way to explore the properties of curves and surfaces in a comprehensive manner. When calculating surface areas like in our method, they form an essential part of transitioning between coordinate systems or implementing advanced calculus operations.

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Most popular questions from this chapter

Work and area Suppose that \(f(t)\) is differentiable and positive for \(a \leq t \leq b .\) Let \(C\) be the path \(\mathbf{r}(t)=t \mathbf{i}+f(t) \mathbf{j}, a \leq t \leq b\) and \(\mathbf{F}=\) yi. Is there any relation between the value of the work integral $$ \int_{C} \mathbf{F} \cdot d \mathbf{r} $$ and the area of the region bounded by the \(t\) -axis, the graph of \(f\) and the lines \(t=a\) and \(t=b ?\) Give reasons for your answer.

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 \(\mathbf{F}=f \nabla g .\) )

Let \(\mathbf{F}_{1}\) and \(\mathbf{F}_{2}\) be differentiable vector fields and let \(a\) and \(b\) be arbitrary real constants. Verify the following identities. a. \(\nabla \cdot\left(a \mathbf{F}_{1}+b \mathbf{F}_{2}\right)=a \nabla \cdot \mathbf{F}_{1}+b \nabla \cdot \mathbf{F}_{2}\) b. \(\nabla \times\left(a \mathbf{F}_{1}+b \mathbf{F}_{2}\right)=a \nabla \times \mathbf{F}_{1}+b \nabla \times \mathbf{F}_{2}\) c. \(\nabla \cdot\left(\mathbf{F}_{1} \times \mathbf{F}_{2}\right)=\mathbf{F}_{2} \cdot \nabla \times \mathbf{F}_{1}-\mathbf{F}_{1} \cdot \nabla \times \mathbf{F}_{2}\)

Outward flux of a gradient field Let \(S\) be the surface of the portion of the solid sphere \(x^{2}+y^{2}+z^{2} \leq a^{2}\) that lies in the first octant and let \(f(x, y, z)=\ln \sqrt{x^{2}+y^{2}+z^{2}} .\) Calculate $$\iint_{S} \nabla f \cdot \mathbf{n} d \sigma$$

Use the Divergence Theorem to find the outward flux of \(\mathbf{F}\) across the boundary of the region \(D\) Thick cylinder \(\quad \mathbf{F}=\ln \left(x^{2}+y^{2}\right) \mathbf{i}-\left(\frac{2 z}{x} \tan ^{-1} \frac{y}{x}\right) \mathbf{j}+\) \(z \sqrt{x^{2}+y^{2}} \mathbf{k}\) D: The thick-walled cylinder \(1 \leq x^{2}+y^{2} \leq 2,-1 \leq z \leq 2\)
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