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Chapter 14: Problem 22

Use a line integral on the boundary to find the area of the following regions. The region bounded by the curve \(\mathbf{r}(t)=\left\langle t\left(1-t^{2}\right), 1-t^{2}\right\rangle,\) for \(-1 \leq t \leq 1\) (Hint: Plot the curve.)

Short Answer

Expert verified
Question: Find the area of the region in the plane bounded by the curve \(\mathbf{r}(t)=\langle t(1-t^2), 1-t^2\rangle\) for \(-1\leq t\leq1\). Answer: The area of the region bounded by the given curve is \(\frac{4}{15}\).

Step by step solution

01

Find the parametric equations of the curve

The given curve is $\mathbf{r}(t)=\langle t(1-t^{2}), 1-t^{2}\rangle$. The parametric equations for this curve are: $$x(t)=t(1-t^2)$$ $$y(t)=1-t^2$$
02

Calculate dy/dt and set the limits

Differentiate \(y(t)\) with respect to \(t\) to find \(\frac{dy}{dt}\): $$\frac{dy}{dt}=-2t$$ We will calculate the line integral only for the right half of the region, where \(0 \leq t \leq 1\).
03

Calculate the line integral

To find the area of the region, we need to calculate the line integral \(\oint_C x\, dy\): $$A=\int_{C} x\, dy = \int_{0}^{1} x(t)\cdot\frac{dy}{dt}\, dt$$ Plug in the expressions of \(x(t)\) and \(\frac{dy}{dt}\) we found earlier: $$A=\int_{0}^{1} t(1-t^2)(-2t)\, dt$$
04

Evaluate the integral

To evaluate the integral, first expand the expression inside the integral: $$A=-2\int_{0}^{1} t^2 - 2t^4\, dt$$ Now evaluate the integral by applying the power rule: $$A=-2\left[\frac{1}{3} t^3 -\frac{1}{5} t^5\right]_0^1$$ $$A=-2\left(\frac{1}{3} - \frac{1}{5}\right)$$
05

Account for symmetry

Since the region is symmetric about the y-axis, the total area A is twice the area of the region on the right side where \(0\leq t\leq1\). So, multiply the result by 2: $$A_{total} = 2A=-2\left(-2\left(\frac{1}{3} - \frac{1}{5}\right)\right)$$ $$A_{total} = \frac{4}{3} - \frac{4}{5}$$ $$A_{total} = \frac{12-8}{15}$$
06

Find the area

Now find the total area of the region: $$A_{total} = \frac{4}{15}$$ The area of the region bounded by the given curve is \(\frac{4}{15}\).

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Key Concepts

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

Parametric Equations
When defining curves, especially in the realm of calculus, we often talk in terms of parametric equations. These equations represent a set of points in space, with each coordinate defined as a function of a third variable, typically denoted as t, known as the parameter. The power of parametric equations lies in their ability to describe motion and curves that are not functions or have complex behavior.

For instance, the given exercise requires finding the area within a curve specified by the parametric equations:
  • \(x(t) = t(1-t^2)\)
  • \(y(t) = 1-t^2\)
By plotting these equations, we can visualize the curve in the plane and observe any symmetries or patterns that may simplify our problem. It's essential to note that since the equations are in parametric form, we can capture more complex relationships between x and y鈥攔elationships that might be impossible to describe with a single, explicit function y=f(x).
Differentiation
To delve deeper into calculus, differentiation is a fundamental process that allows us to find the rate at which a function changes at any given point. It's the cornerstone of studying motion, change, and optimization problems. In the context of the exercise, differentiation is used to calculate a derivative, which is essential when plugging into the line integral formula.

Here's how differentiation is applied to our problem:
  • We have y(t), and we need its derivative dy/dt.
  • By applying the rules of differentiation, specifically the power rule, we obtain \(\frac{dy}{dt} = -2t\).
This derivative represents the instantaneous rate of change of y with respect to the parameter t. Having this information is crucial because when we compute the line integral to find the area, we'll be integrating x(t) multiplied by this derivative.
Power Rule Integral
The power rule integral is one of the most basic yet powerful tools in calculus for finding antiderivatives of monomial functions. It's essentially the reverse operation of the power rule for differentiation.

To apply the power rule for integration, which is necessary in the final steps of our exercise solution, we follow the pattern:
  • For a term like \(t^n\), the antiderivative is \(\frac{t^{n+1}}{n+1}\), if \(neq -1\).
In our solution, we use the power rule to integrate \(t^2\) and \(t^4\), getting \(\frac{1}{3}t^3\) and \(\frac{1}{5}t^5\) respectively as antiderivatives, leading us to the area of the given region.

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Most popular questions from this chapter

Recall the Product Rule of Theorem \(14.11: \nabla \cdot(u \mathbf{F})=\nabla u \cdot \mathbf{F}+u(\nabla \cdot \mathbf{F})\) a. Integrate both sides of this identity over a solid region \(D\) with a closed boundary \(S\) and use the Divergence Theorem to prove an integration by parts rule: $$\iiint_{D} u(\nabla \cdot \mathbf{F}) d V=\iint_{S} u \mathbf{F} \cdot \mathbf{n} d S-\iiint_{D} \nabla u \cdot \mathbf{F} d V$$ b. Explain the correspondence between this rule and the integration by parts rule for single-variable functions. c. Use integration by parts to evaluate \(\iiint_{D}\left(x^{2} y+y^{2} z+z^{2} x\right) d V\) where \(D\) is the cube in the first octant cut by the planes \(x=1\) \(y=1,\) and \(z=1\)

Begin with the paraboloid \(z=x^{2}+y^{2},\) for \(0 \leq z \leq 4,\) and slice it with the plane \(y=0\) Let \(S\) be the surface that remains for \(y \geq 0\) (including the planar surface in the \(x z\) -plane) (see figure). Let \(C\) be the semicircle and line segment that bound the cap of \(S\) in the plane \(z=4\) with counterclockwise orientation. Let \(\mathbf{F}=\langle 2 z+y, 2 x+z, 2 y+x\rangle\) a. Describe the direction of the vectors normal to the surface that are consistent with the orientation of \(C\). b. Evaluate \(\iint_{S}(\nabla \times \mathbf{F}) \cdot \mathbf{n} d S\) c. Evaluate \(\oint_{C} \mathbf{F} \cdot d \mathbf{r}\) and check for agreement with part (b).

Prove the following properties of the divergence and curl. Assume \(\mathbf{F}\) and \(\mathbf{G}\) are differentiable vector fields and \(c\) is a real number. a. \(\nabla \cdot(\mathbf{F}+\mathbf{G})=\nabla \cdot \mathbf{F}+\nabla \cdot \mathbf{G}\) b. \(\nabla \times(\mathbf{F}+\mathbf{G})=(\nabla \times \mathbf{F})+(\nabla \times \mathbf{G})\) c. \(\nabla \cdot(c \mathbf{F})=c(\nabla \cdot \mathbf{F})\) d. \(\nabla \times(c \mathbf{F})=c(\nabla \times \mathbf{F})\)

Let S be the disk enclosed by the curve \(C: \mathbf{r}(t)=\langle\cos \varphi \cos t, \sin t, \sin \varphi \cos t\rangle,\)for \(0 \leq t \leq 2 \pi,\) where \(0 \leq \varphi \leq \pi / 2\) is a fixed angle. What is the circulation on \(C\) of the vector field \(\mathbf{F}=\langle-y,-z, x\rangle\) as a function of \(\varphi ?\) For what value of \(\varphi\) is the circulation a maximum?

A square plate \(R=\\{(x, y): 0 \leq x \leq 1,\) \(0 \leq y \leq 1\\}\) has a temperature distribution \(T(x, y)=100-50 x-25 y\) a. Sketch two level curves of the temperature in the plate. b. Find the gradient of the temperature \(\nabla T(x, y)\) c. Assume that the flow of heat is given by the vector field \(\mathbf{F}=-\nabla T(x, y) .\) Compute \(\mathbf{F}\) d. Find the outward heat flux across the boundary \(\\{(x, y): x=1,0 \leq y \leq 1\\}\) e. Find the outward heat flux across the boundary \(\\{(x, y): 0 \leq x \leq 1, y=1\\}\)
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