Question

Test for independence of path and evaluate the following integrals: a) \(\int_{(1,-2)}^{(3,4)} \frac{y d x-x d y}{x^{2}}\) on the line \(y=3 x-5\); b) \(\int_{(0,2)}^{(1,3)} \frac{3 x^{2}}{y} d x-\frac{x^{3}}{y^{2}} d y\) on the parabola \(y=2+x^{2}\). c) \(\int_{(1,0)}^{(-1,0)}(2 x y-1) d x+\left(x^{2}+6 y\right) d y\) on the circular arc \(y=\sqrt{1-x^{2}},-1 \leq x \leq 1\); d) \(\int_{(0,0)}^{\left(\frac{\pi}{4}, \frac{\pi}{4}\right)} \sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y\) on the curve \(y=16 x^{3} / \pi^{2}\).

Short answer

#tag_title# Evaluate the Integral - Part b (Continued) #tag_content# To solve the remaining integral: \(\int_0^1 \frac{2x^4}{2+x^2} dx\) We perform the substitution: \(u = x^2 \Rightarrow du = 2xdx\), then: \(\begin{aligned} \int_0^1 \frac{2x^4}{2+x^2} dx &= \int \frac{u^2}{2+u} du \\ &= \int (\frac{u^2-2u+2}{2+u} + 2u - 2) du \end{aligned}\) This simplification was done using partial fractions. Now, we can evaluate the integral: \(\begin{aligned} \int_0^1 (\frac{u^2-2u+2}{2+u} + 2u - 2)du &= [2(u-\ln|u+2|- 2) - 2\ln|2+u|]_0^1 \\ &= (2(1-\ln(3)) - 2\ln(3)) - (0) \\ &= 2-2\ln(3) \end{aligned}\) Now, we can finalize the answer for part b: \(1^3 - \int_0^1 \frac{2x^4}{2+x^2} dx = 1 - (2 - 2\ln(3)) = -1 + 2\ln(3)\) #tag_title# Evaluate the Integral - Part c (Continued) #tag_content# To evaluate the remaining integral in part c, we will use trigonometric substitution. Let \(x = \sin(\theta)\), so \(dx = \cos(\theta)d\theta\): \(\begin{aligned} \int_{1}^{-1} (2x\sqrt{1-x^2}-1)dx - \left(\int_{1}^{-1} (x^3 -\frac{6x^2}{1-x^2}) dx\right) &= \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (2\sin(\theta)\cos^2(\theta)-\cos(\theta))d\theta - \left(-\int_{-1}^1 6x^2dx\right) \end{aligned}\) Now, we can evaluate the integrals: \(\begin{aligned} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (2\sin(\theta)\cos^2(\theta)-\cos(\theta))d\theta &= [\sin^3(\theta)]_{\frac{\pi}{2}}^{\frac{3\pi}{2}} + \int_\frac{\pi}{2}^\frac{3\pi}{2}\cos(\theta)d\theta \\ &= -2 + [(\sin(\theta)]_\frac{\pi}{2}^\frac{3\pi}{2} \\ &= -2 + (-1-1) \\ &= -4 \end{aligned}\) \(\int_{-1}^1 6x^2dx = [2x^3]_{-1}^1 = (2 - (-2)) = 4\) Adding the two values, we get the final answer for part c: \((-4) - 4 = -8\) The answers for parts b and c are: b) \(-1 + 2\ln(3)\) c) \(-8\)

Step-by-step solution

Test for Path Independence - Part a

To check if \(\int \frac{y dx - x dy}{x^2}\) is path-independent, we need to see if the given expression forms a conservative vector field. For that, we can check if the given integrand \(\frac{y}{x^2} dx- \frac{x}{x^2} dy\) has a potential function. Taking the partial derivatives, we get: \(\frac{\partial}{\partial y}\left(\frac{y}{x^2}\right)=-\frac{\partial}{\partial x}\left(\frac{x}{x^2}\right)\) \(\frac{1}{x^2}=-\frac{-2x}{x^3}\) \(1=-2\) Since the condition is not satisfied, the vector field is not conservative. Therefore, it is path-dependent, and we cannot evaluate the given integral.

Test for Path Independence - Part b

For this integral, \(\int \frac{3x^2}{y} dx - \frac{x^3}{y^2} dy\), let's follow the same procedure: \(\frac{\partial}{\partial y}\left(\frac{3x^2}{y}\right)=-\frac{\partial}{\partial x}\left(\frac{x^3}{y^2}\right)\) \(\frac{-3x^2}{y^2}=-\frac{3x^2}{y^2}\) The condition is satisfied, so the integral is path-independent.

Test for Path Independence - Part c

Now, let's test the integral \(\int (2xy-1)dx +(x^2+6y)dy\) for path independence: \(\frac{\partial}{\partial y}(2xy-1)=-\frac{\partial}{\partial x}(x^2+6y)\) \(2x=-2x\) The condition is satisfied, so the integral is path-independent.

Test for Path Independence - Part d

For the integral \(\int \sec^2 x \tan y\, dx + \sec^2 y \tan x\, dy\), let's perform the path independence test: \(\frac{\partial}{\partial y}(\sec^2 x \tan y)=-\frac{\partial}{\partial x}(\sec^2 y \tan x)\) \(\sec^2 x \sec^2 y = -\sec^2 y \sec^2 x\) The given condition is not satisfied; therefore, the integral is path-dependent and cannot be evaluated. Since we can only evaluate path-independent integrals, we will proceed with the evaluations of parts b and c. #Step 2: Evaluate path-independent integrals#

Evaluate the Integral - Part b

To evaluate the given integral \(\int_{(0,2)}^{(1,3)} \frac{3x^2}{y} dx - \frac{x^3}{y^2} dy\) on the parabola \(y=2+x^2\), let's first rewrite the integrand in terms of a single variable. Using the parabola equation, we get \(dy = 2xdx\), so we can rewrite the given integral as: \(\int_0^1 \frac{3x^2}{2+x^2}dx - \frac{x^3}{(2+x^2)^2}(2xdx)\) Now we can combine the integrals and simplify the expression: \(\int_0^1 (3x^2-\frac{2x^4}{2+x^2}) dx\) Now, evaluate the integral: \(\begin{aligned} \int_0^1 (3x^2-\frac{2x^4}{2+x^2}) dx &= \left[ \frac{3}{3}x^3 - \int \frac{2x^4}{2+x^2} dx \right]_0^1 \\ &= \left[ x^3 - \int \frac{2x^4}{2+x^2} dx \right]_0^1 \\ &= 1^3 - \int_0^1 \frac{2x^4}{2+x^2} dx \\ \end{aligned}\) To solve the remaining integral, you may use integration techniques such as substitution or partial fractions. The final result should be the answer to part b.

Evaluate the Integral - Part c

To evaluate the given integral \(\int_{(1,0)}^{(-1,0)}(2xy-1)dx +(x^2+6y)dy\) on the circular arc \(y=\sqrt{1-x^2}\), let's first rewrite the integral in terms of a single variable (x in this case). Using the equation of the circle, we get \(dy = -\frac{x}{\sqrt{1-x^2}}dx\), so we can rewrite the given integral as: \(\int_{1}^{-1} (2x(\sqrt{1-x^2})-1)dx -\left((x^2+\frac{6x}{\sqrt{1-x^2}})(-\frac{x}{\sqrt{1-x^2}} dx)\right)\) Now we can simplify the expression and evaluate the integral: \(\int_{1}^{-1} (2x\sqrt{1-x^2}-1)dx - \left(\int_{1}^{-1} (x^3 -\frac{6x^2}{1-x^2}) dx\right)\) The final result should be the answer to part c. You may use integration techniques such as substitution or trigonometric substitution to solve the remaining integral.

Key concepts

Path Independence

The concept of path independence relates to whether a line integral depends solely on its start and end points, or if the particular path taken affects the outcome. When a line integral is path-independent, it suggests that the vector field is conservative. This means that there exists a scalar potential function whose gradient matches the vector field.

To test for path independence, check if the curl of the vector field is zero. If it is, the vector field is conservative, and hence, the line integral is path-independent. In our exercise, parts b and c are path-independent as their conditions are satisfied. Parts a and d do not meet these conditions, indicating path dependence, so they cannot be easily evaluated.

Understanding path independence is crucial in fields like physics and engineering, as it provides simplicity in calculating work done by conservative forces.

Conservative Vector Field

A conservative vector field is one where the line integral between two points is the same for all paths connecting those points, expressing a key feature of physical systems like gravitational or electrostatic fields.

In mathematical terms, a conservative vector field has a potential function, such that the gradient of this potential equals the vector field itself.

To determine if a vector field is conservative, verify that the mixed partial derivatives are continuous and equal, indicating path independence and a potential function exists. In our exercise, parts b and c satisfy this condition.

Conservative fields have profound implications in simplifying complex systems, making them easier to analyze and manipulate.

Evaluation of Integrals

Evaluating line integrals involves integrating a function over a path, taking into account the direction and length of the path.

In our exercise, we evaluated the integral for part b by substituting the path equation into the functions, transforming the problem into a single-variable integral. Similarly, for part c, the circular path allowed for simplification through trigonometric substitution.

Here are the key steps:

• Substitute the path equation into the differential expressions.
• Simplify the integrand to a manageable form.
• Evaluate the simplified integral using appropriate methods, like substitution or partial fractions.

Mastering these techniques is essential for solving complex physical and mathematical problems effectively.

Arc Length Parameterization

Arc length parameterization is a method for representing curves so that each segment along the curve corresponds to an equal proportion of the total arc length. This simplifies the computation of integrals on curves since the path can be described succinctly using a parameter.

In part c of our exercise, the path was a circular arc, which was expressed using the arc length parameterization. This allowed us to describe the curve with respect to the parameter, enabling easier integration.

Key benefits include:

• Enables uniform scaling along the curve.
• Facilitates integration by providing a natural parameterization.
• Allows accurate calculation of quantities like arc length and curvature.

This technique is widely adopted in physics and engineering for its efficiency in dealing with curves and paths.