Problem 51 - Calculus Early Transcendentals

Question

Evaluate the following integrals using the method of your choice. A sketch is helpful. $$\int_{-1}^{1} \int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\left(x^{2}+y^{2}\right)^{3 / 2} d y d x$$

Step-by-Step Solution

Answer

Question: Evaluate the double integral $$\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (x^2+y^2)^{3/2} dy dx$$ Answer: The double integral evaluates to $$\frac{\pi}{3}$$

Step 1: Change into Polar Coordinates

To convert the given integral into polar coordinates, we use the substitutions \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). Also, when we change coordinates, we need to multiply by the Jacobian determinant, which is \(r\) for polar coordinates. Our integral becomes: $$\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (x^2+y^2)^{3/2} dy dx \Rightarrow \int_{0}^{1} \int_{0}^{2\pi} (r^2)^{3/2} r d\theta dr$$

Step 2: Evaluate the \(\theta\) integral

Now we evaluate the integral with respect to \(\theta\), while keeping \(r\) as a constant: $$\int_{0}^{2\pi} d\theta = 2\pi$$

Step 3: Simplify the \(r\) integral

Now let's simplify the \(r\) integral involving \((r^2)^{3/2}\) and an extra factor of \(r\): $$\int_{0}^{1} (r^2)^{3/2} r dr = \int_{0}^{1} r^5 dr$$

Step 4: Evaluate the \(r\) integral

Next, integrate with respect to \(r\): $$\int_{0}^{1} r^5 dr = \left[\frac{r^6}{6}\right]_{0}^{1} = \frac{1^6}{6} - \frac{0^6}{6} = \frac{1}{6}$$

Step 5: Combine the results

Now, combine the results from evaluating the \(\theta\) integral and the \(r\) integral to get the final answer: $$\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (x^2+y^2)^{3/2} dy dx = 2\pi \times \frac{1}{6} = \boxed{\frac{\pi}{3}}$$

Key Concepts

Integral Calculus

Integral calculus is all about finding the accumulation of quantities. Think of it as a way to calculate areas under curves or finding the total amount of something. In simple terms, when we have a function and we want to find out how it behaves over an interval, we integrate it. There are two main types of integrals: definite and indefinite. - **Definite Integrals**: These have upper and lower limits and give a numerical result. They are used to find the area under a curve over an interval, like our exercise with limits from \(-1\) to \(1\).- **Indefinite Integrals**: These do not have specified limits and give a general formula or function.In our exercise, we're using a double integral, which is just multiple integrals done together to find the volume under a surface. The integral starts in Cartesian coordinates, but it's easier to convert it into polar coordinates, especially when dealing with circular shapes.

Double Integrals

A double integral extends the concept of integration to functions of two variables. It's like stacking slices of areas to find a volume. When you evaluate a double integral:1. **Setup**: Identify the limits of integration. For instance, our exercise started with limits in Cartesian form.2. **Evaluate**: Usually by breaking it into two steps: first integrate with respect to one variable, then the other.Switching to polar coordinates in the exercise makes evaluation straightforward:- **Polar Coordinates**: Represent points using radius \(r\) and angle \(\theta\).- **In Polar Form**: We integrate first over \(\theta\) from \(0\) to \(2\pi\), representing a full circle, and then over \(r\) from \(0\) to \(1\), representing the radius.Using polar coordinates simplifies dealing with circular regions, converting complex cartesian limits into more manageable forms.

Jacobian Determinant

The Jacobian determinant is crucial when changing variables in multiple integrals. It adjusts for the change in area or volume caused by the new coordinate system. - **Why Use It?**: When converting from Cartesian \( (x, y) \) to Polar \( (r, \theta) \) coordinates, areas and shapes stretch or shrink. The Jacobian helps correct this.- **For Polar Coordinates**: The Jacobian determinant is \(r\). That's why when our exercise changed from Cartesian to polar coordinates, we multiplied the integrand by \(r\).Here's what happens:- The transformation formulas \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\) are used.- The Jacobian determinant compensates for how much the new variables distort the original region.This makes sure our integral in the new coordinates gives the same result as if it were in the original coordinates.