Problem 22 - Thomas Calculus

Question

In Exercises \(9-22,\) change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral. $$ \int_{0}^{1} \int_{x}^{\sqrt{2-x^{2}}}(x+2 y) d y d x $$

Step-by-Step Solution

Answer

The value of the polar integral is 2.

Step 1: Understanding Cartesian Limits

First, analyze the given Cartesian integral limits \(x\) and \(y\). The integral \(\int_{0}^{1} \int_{x}^{\sqrt{2-x^{2}}}(x+2y)\, dy\, dx\) indicates that \(x\) ranges from 0 to 1 and for each \(x\), \(y\) ranges from \(x\) to \(\sqrt{2-x^2}\).

Step 2: Conversion to Polar Coordinates

In polar coordinates, \(x = r \cos\theta\) and \(y = r\sin\theta\). The expression \((x+2y)\) becomes \((r\cos\theta + 2r\sin\theta)\). Also, noting that \(\sqrt{2 - x^2}\) in terms of \(r\) and \(\theta\) becomes \(r = \sqrt{2}\).

Step 3: Determine Integral Limits in Polar Coordinates

The Cartesian limits \(0 \leq x \leq 1\) and \(x \leq y \leq \sqrt{2-x^2}\) suggest the region is constrained by the quarter circle from \(\theta = 0\) to \(\theta = \pi/4\) and \(r\) from \(0\) to \(\sqrt{2}\).

Step 4: Set Up the Polar Integral

The Jacobian of the transformation from Cartesian to polar coordinates introduces an additional \(r\) term, meaning the differential element changes from \(dx\,dy\) to \(r\,dr\,d\theta\). Thus, the polar integral becomes \[ \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sqrt{2}} (r\cos\theta + 2r\sin\theta) r\, dr\, d\theta. \]

Step 5: Evaluate the Inner Integral

Evaluate the inner integral with respect to \(r\):\[ \int_{0}^{\sqrt{2}} (r^2\cos\theta + 2r^2\sin\theta)\, dr = \left[ \frac{r^3\cos\theta}{3} + \frac{2r^3\sin\theta}{3} \right]_{0}^{\sqrt{2}} = \frac{2\sqrt{2}}{3}(\cos\theta + 2\sin\theta). \]

Step 6: Evaluate the Outer Integral

Now integrate with respect to \(\theta\) from 0 to \(\frac{\pi}{4}\):\[ \int_{0}^{\frac{\pi}{4}} \frac{2\sqrt{2}}{3} (\cos\theta + 2\sin\theta)\, d\theta. \] Split it into two integrals:1. \( \int_{0}^{\frac{\pi}{4}} \cos\theta\, d\theta \)2. \( 2\int_{0}^{\frac{\pi}{4}} \sin\theta\, d\theta \).

Step 7: Evaluate Each Separate Integral

For \( \int \cos\theta\, d\theta \), the integration gives \(\sin\theta\), therefore from 0 to \(\frac{\pi}{4}\), it becomes \(\sqrt{2}/2.\). For \( 2\int \sin\theta\, d\theta \), the integration yields \(-2\cos\theta\), evaluated from 0 to \(\frac{\pi}{4}\) gives \(2\sqrt{2}/2 = \sqrt{2}.\).

Step 8: Calculate Final Result

Combine the results to complete the integration:\[ \frac{2\sqrt{2}}{3} \left( \frac{\sqrt{2}}{2} + \sqrt{2} \right) = \frac{2\sqrt{2}}{3} \cdot \frac{3\sqrt{2}}{2} = 2. \]

Key Concepts

Cartesian to Polar Conversion

Imagine moving from one kind of map to another. That's what happens when we convert Cartesian coordinates to polar coordinates. In Cartesian coordinates, we define points using a grid system with an \(x\)-axis and a \(y\)-axis. However, in polar coordinates, we describe points using an angle and a distance from the origin.

To make these conversions:

• The **x-coordinate** becomes \(x = r \cos \theta\).
• The **y-coordinate** becomes \(y = r \sin \theta\).

Here, \(r\) is the distance from the origin to the point, and \(\theta\) is the angle made with the positive \(x\)-axis. Converting equations from one form to another can simplify problems, especially in calculus, where integrating over curved areas is easier using polar coordinates.

Integral Calculus

Integral calculus is the study of accumulations, such as areas under curves. It allows us to find the total amount of something if we know the rate of accumulation. For instance, in this exercise, we are asked to integrate a function over a specific region in the \(x, y\) plane. This process involves two simple steps:

• First, decide the limits of integration, which define the area over which we integrate.
• Second, evaluate the integrals, which can be considered like summing up small parts of the area to find the whole.

The conversion to polar coordinates often simplifies calculating these integrals, as polar limits can be more natural when dealing with curves and circles. Integral calculus is a fundamental tool which allows mathematicians, scientists, and engineers to solve real-world problems involving continuous change.

Jacobian Transformation

When transforming coordinates from Cartesian to polar, the area element \(dx\,dy\) gets transformed too. This is where the Jacobian comes in handy. It is the factor by which we need to multiply to switch to polar's area element \(r\,dr\,d\theta\).

The Jacobian transformation is essential for ensuring that the integral’s value remains unchanged even after the transformation. Mathematically, it is represented as:

• For conversion from Cartesian to polar, the Jacobian is \(r\), meaning that you multiply the function by \(r\) when setting up the polar integral.

Understanding this concept helps in making sure our integration covers the same physical area before and after transformation and the integral calculation stays accurate.

Polar Integral Evaluation

Evaluating polar integrals involves careful consideration once we have converted the equations and limits. The given exercise shows the setup of a new integral in polar coordinates:

• First, substitute the polar forms and Jacobian into the integral’s expression.
• Next, solve the **inner integral** with respect to \(r\).
• Finally, solve the **outer integral** with respect to \(\theta\).

Through this procedure, we integrate in a region curved due to the circle's equation, simplifying the steps compared to rectangular coordinates. This exercise demonstrates that sometimes, adopting an entirely different perspective on the same problem can lead to simpler solutions. Polar coordinates and the subsequent integral evaluation reflect this beautifully, leading to an elegant solution in this instance.