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Chapter 13: Problem 51

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

Short Answer

Expert verified
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 by step solution

01

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: $$\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (x^2+y^2)^{3/2} dy dx \Rightarrow \int_{0}^{1} \int_{0}^{2\pi} (r^2)^{3/2} r d\theta dr$$
02

Evaluate the \(\theta\) integral

Now we evaluate the integral with respect to \(\theta\), while keeping \(r\) as a constant: $$\int_{0}^{2\pi} d\theta = 2\pi$$
03

Simplify the \(r\) integral

Now let's simplify the \(r\) integral involving \((r^2)^{3/2}\) and an extra factor of \(r\): $$\int_{0}^{1} (r^2)^{3/2} r dr = \int_{0}^{1} r^5 dr$$
04

Evaluate the \(r\) integral

Next, integrate with respect to \(r\): $$\int_{0}^{1} r^5 dr = \left[\frac{r^6}{6}\right]_{0}^{1} = \frac{1^6}{6} - \frac{0^6}{6} = \frac{1}{6}$$
05

Combine the results

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

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Key Concepts

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

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.

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Most popular questions from this chapter

The occurrence of random events (such as phone calls or e-mail messages) is often idealized using an exponential distribution. If \(\lambda\) is the average rate of occurrence of such an event, assumed to be constant over time, then the average time between occurrences is \(\lambda^{-1}\) (for example, if phone calls arrive at a rate of \(\lambda=2 /\) min, then the mean time between phone calls is \(\lambda^{-1}=\frac{1}{2} \mathrm{min}\) ). The exponential distribution is given by \(f(t)=\lambda e^{-\lambda t},\) for \(0 \leq t<\infty\) a. Suppose you work at a customer service desk and phone calls arrive at an average rate of \(\lambda_{1}=0.8 /\) min (meaning the average time between phone calls is \(1 / 0.8=1.25 \mathrm{min}\) ). The probability that a phone call arrives during the interval \([0, T]\) is \(p(T)=\int_{0}^{T} \lambda_{1} e^{-\lambda_{1} t} d t .\) Find the probability that a phone call arrives during the first 45 s \((0.75\) min) that you work at the desk. b. Now suppose that walk-in customers also arrive at your desk at an average rate of \(\lambda_{2}=0.1 /\) min. The probability that a phone $$p(T)=\int_{0}^{T} \int_{0}^{T} \lambda_{1} e^{-\lambda_{1} t} \lambda_{2} e^{-\lambda_{2} x} d t d s$$ Find the probability that a phone call and a customer arrive during the first 45 s that you work at the desk. c. E-mail messages also arrive at your desk at an average rate of \(\lambda_{3}=0.05 /\) min. The probability that a phone call and a customer and an e-mail message arrive during the interval \([0, T]\) is $$p(T)=\int_{0}^{T} \int_{0}^{T} \int_{0}^{T} \lambda_{1} e^{-\lambda_{1} t} \lambda_{2} e^{-\lambda_{2} s} \lambda_{3} e^{-\lambda_{3} u} d t d s d u$$ Find the probability that a phone call and a customer and an e-mail message arrive during the first 45 s that you work at the desk.

Use a change of variables to evaluate the following integrals. \(\iiint_{D} d V ; D\) is bounded by the upper half of the ellipsoid \(x^{2} / 9+y^{2} / 4+z^{2}=1\) and the \(x y\) -plane. Use \(x=3 u\) \(y=2 v, z=w\)

Let \(D\) be the solid bounded by the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1,\) where \(a>0, b>0,\) and \(c>0\) are real numbers. Let \(T\) be the transformation \(x=\)au, \(y=b v, z=c w\) Find the average square of the distance between points of \(D\) and the origin.

Limiting center of mass \(A\) thin rod of length \(L\) has a linear density given by \(\rho(x)=\frac{10}{1+x^{2}}\) on the interval \(0 \leq x \leq L\). Find the mass and center of mass of the rod. How does the center of mass change as \(L \rightarrow \infty ?\)

Open and closed boxes Consider the region \(R\) bounded by three pairs of parallel planes: \(a x+b y=0, a x+b y=1\) \(c x+d z=0, c x+d z=1, e y+f z=0,\) and \(e y+f z=1\) where \(a, b, c, d, e,\) and \(f\) are real numbers. For the purposes of evaluating triple integrals, when do these six planes bound a finite region? Carry out the following steps. a. Find three vectors \(\mathbf{n}_{1}, \mathbf{n}_{2},\) and \(\mathbf{n}_{3}\) each of which is normal to one of the three pairs of planes. b. Show that the three normal vectors lie in a plane if their triple scalar product \(\mathbf{n}_{1} \cdot\left(\mathbf{n}_{2} \times \mathbf{n}_{3}\right)\) is zero. c. Show that the three normal vectors lie in a plane if ade \(+b c f=0\) d. Assuming \(\mathbf{n}_{1}, \mathbf{n}_{2},\) and \(\mathbf{n}_{3}\) lie in a plane \(P,\) find a vector \(\mathbf{N}\) that is normal to \(P .\) Explain why a line in the direction of \(\mathbf{N}\) does not intersect any of the six planes and therefore the six planes do not form a bounded region. e. Consider the change of variables \(u=a x+b y, v=c x+d z\) \(w=e y+f z .\) Show that $$J(x, y, z)=\frac{\partial(u, v, w)}{\partial(x, y, z)}=-a d e-b c f$$ What is the value of the Jacobian if \(R\) is unbounded?
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