Problem 29 - Thomas Calculus

Question

In Exercises \(27 - 30\) , integrate \(f\) over the given curve. $$ \begin{array} { l } { f ( x , y ) = x + y , \quad C : \quad x ^ { 2 } + y ^ { 2 } = 4 \text { in the first quadrant from } } \\ { ( 2,0 ) \text { to } ( 0,2 ) } \end{array} $$

Step-by-Step Solution

Answer

The integral evaluates to 8.

Step 1: Understand the Curve

The curve given is a circle centered at the origin with radius 2, but we only consider the first quadrant from the point (2,0) to (0,2). This part of the circle can be represented parametrically by setting \(x = 2 \cos(t)\) and \(y = 2 \sin(t)\) for \(t\) ranging from \(0\) to \(\frac{\pi}{2}\).

Step 2: Parameterize the Curve

Using the parametric equations given by \(x = 2 \cos(t)\) and \(y = 2 \sin(t)\), we can express the differential arc length \(ds\) by first differentiating: \(\frac{dx}{dt} = -2 \sin(t)\) and \(\frac{dy}{dt} = 2 \cos(t)\). Therefore, \(ds = \sqrt{(-2 \sin(t))^2 + (2 \cos(t))^2} dt = 2 dt\).

Step 3: Set Up the Integral

The line integral of \(f(x, y) = x + y\) over the curve is expressed in terms of \(t\) and \(ds\). Substitute \(x = 2 \cos(t)\) and \(y = 2 \sin(t)\) into the function: \(f(x, y) = 2\cos(t) + 2\sin(t)\). Multiply by \(2\) (from \(ds = 2 dt\)) to set up the integral: \(\int_{0}^{\frac{\pi}{2}} (2\cos(t) + 2\sin(t))*2 \, dt\).

Step 4: Solve the Integral

Calculate the integral \(\int_{0}^{\frac{\pi}{2}} (4\cos(t) + 4\sin(t)) \, dt\). Separate into two integrals: \(4\int_{0}^{\frac{\pi}{2}} \cos(t) \, dt + 4\int_{0}^{\frac{\pi}{2}} \sin(t) \, dt\). These integrate to \(4 [\sin(t)]_{0}^{\frac{\pi}{2}} + 4 [- \cos(t)]_{0}^{\frac{\pi}{2}}\).

Step 5: Evaluate the Integrals and Simplify

Evaluate each part: \(4(\sin(\frac{\pi}{2}) - \sin(0)) = 4(1 - 0) = 4\) and \(4(-\cos(\frac{\pi}{2}) + \cos(0)) = 4(0 + 1) = 4\). Combine these results to get \(4 + 4 = 8\). Thus, the integral evaluates to 8.

Key Concepts

Parametric Equations

Parametric equations are a way to represent curves using a parameter, often labeled as \(t\). In our problem, we're dealing with a segment of a circle, so using parametric equations is very helpful. For a circle centered at the origin with radius \(2\), we can describe it in two dimensions.

• The **x-coordinate** is expressed as \(x = 2 \cos(t)\).
• The **y-coordinate** is represented by \(y = 2 \sin(t)\).

These equations are valid as \(t\) varies from \(0\) to \(\frac{\pi}{2}\), effectively defining the segment of the circle in the first quadrant. This approach helps us manage both coordinates simultaneously and is particularly useful for evaluating integrals over curves, like the line integrals in our exercise.

Arc Length

The concept of arc length is crucial for understanding how to measure the length of a curve defined by parametric equations. In our exercise, we specifically need to find the differential arc length, \(ds\), which allows us to set up the integral. To do this:

• Differentiating **x** gives: \(\frac{dx}{dt} = -2 \sin(t)\).
• Differentiating **y** results in: \(\frac{dy}{dt} = 2 \cos(t)\).

Using these derivatives, we can calculate \(ds\) as:\[ ds = \sqrt{(-2 \sin(t))^2 + (2 \cos(t))^2} \, dt = 2 \, dt\]The differential arc length shows that each infinitesimal piece of the curve, when traversing the arc from \(t = 0\) to \(t = \frac{\pi}{2}\), is equivalent to a simple \(2 \, dt\). This turns our line integral into a more straightforward computation!

Trigonometric Integration

Trigonometric integration involves integrating functions that include trigonometric functions, such as sine and cosine. In our problem, we're dealing with an integral of a function \(f(x, y)\), where \(x\) and \(y\) are replaced using parametric equations:

• Substitute to get: \(f(x, y) = 2 \cos(t) + 2 \sin(t)\).
• Remember the differential arc length: \(ds = 2 \, dt\).

This transforms the line integral into solving:\[ \int_{0}^{\frac{\pi}{2}} (4 \cos(t) + 4 \sin(t)) \, dt\]To solve, you simply separate into two integrals:

• \(4 \int_{0}^{\frac{\pi}{2}} \cos(t) \, dt\)
• \(+ 4 \int_{0}^{\frac{\pi}{2}} \sin(t) \, dt\)

Upon evaluation:

• The cosine integral yields \(4 (\sin(t))\) evaluated from \(0\) to \(\frac{\pi}{2}\), resulting in \(4\).
• The sine integral results in \(4 (-\cos(t))\), also evaluating to \(4\).

Combining these results gives an overall integral value of \(8\). Trigonometric integrals, like those with sine and cosine, often appear in calculus and can be efficiently solved using substitution and standard antiderivatives.