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Cartesian to polar coordinates Sketch the given region of integration \(R\) and evaluate the integral over \(R\) using polar coordinates. $$\iint_{R} e^{-x^{2}-y^{2}} d A ; R=\left\\{(x, y): x^{2}+y^{2} \leq 9\right\\}$$

Short Answer

Expert verified
Question: Evaluate the integral \(\iint_R e^{-x^2-y^2} dA\) over the region \(R = \{(x, y): x^2+y^2 \leq 9\}\) in polar coordinates. Answer: \((1 - e^{-9})\pi\)

Step by step solution

01

Sketch the region of integration

The given region of integration is: \(R = \{(x, y): x^2+y^2 \leq 9\}\). This represents a disk with center at the origin (0, 0) and radius 3. To sketch this region, draw a circle with center at the origin and radius 3, and shade the area inside the circle.
02

Convert to polar coordinates

We will now convert the given limits of integration and integrand to polar coordinates. Recall that in polar coordinates, \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). Therefore, \(x^2 + y^2 = r^2\). The given region of integration becomes: $$R = \{(r, \theta): 0 \leq r \leq 3, 0 \leq \theta \leq 2\pi\}$$ The integrand in the given integral is \(e^{-x^2-y^2}\), which can be rewritten in polar coordinates as \(e^{-r^2}\). We also need to account for the Jacobian of the transformation from Cartesian to polar coordinates, which is given by the absolute value of the determinant of the transformation matrix. In this case, the Jacobian is equal to \(r\). Now, we can rewrite the integral in polar coordinates as: $$\iint_R e^{-r^2}r dr d\theta$$
03

Evaluate the integral

Now, evaluate the integral with the new limits of integration and integrand in polar coordinates: $$\int_{0}^{2\pi} \int_{0}^{3} e^{-r^2}r dr d\theta$$ First, integrate with respect to r: $$\int_{0}^{2\pi} \left[-\frac{1}{2} e^{-r^2}\right]_0^3 d\theta$$ The term \(e^{-r^2}\) is integrated as \(-\frac{1}{2}e^{-r^2}\), which when evaluated at the limits, becomes: $$\int_{0}^{2\pi} \left[\frac{1}{2} -\frac{1}{2} e^{-9}\right] d\theta$$ Now, integrate with respect to \(\theta\): $$\left[\frac{1}{2}\theta - \frac{1}{2}e^{-9}\theta\right]_0^{2\pi}$$ Evaluating the integral at the limits gives: $$\left(\frac{1}{2}(2\pi) - \frac{1}{2}e^{-9}(2\pi)\right) - \left(0 - 0\right) = \left(1 - e^{-9}\right)\pi$$ So, the value of the integral over the region R in polar coordinates is: $$\boxed{(1 - e^{-9})\pi}$$

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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 a branch of mathematics that focuses on finding the total quantity by summing infinitesimally small amounts. In simpler terms, it helps us calculate the area under a curve or the region defined by some boundaries.

In this exercise, we're asked to evaluate the integral of the function \( e^{-x^2-y^2} \) over a circular region in the xy-plane with a radius of 3. This integral represents the accumulated effect or the area under the surface described by our function within this circular boundary.

When transitioning into polar coordinates, one significant advantage is simplifying integration over circular regions. This is achieved by transforming the problem into a different coordinate system (polar) that aligns more naturally with the shape of the region of integration.
Coordinate Transformation
Coordinate transformation involves changing from one system of coordinates to another. It is often used when a different coordinate system simplifies the problem or better fits the nature of a problem's domain.

In this exercise, we change from Cartesian coordinates, where a point \((x, y)\) lies on a flat, square grid, to polar coordinates, where a point is defined by a radius \(r\) and an angle \(\theta\). This makes doing calculations more straightforward in regions like circles or curves.

Some equations become easier in polar coordinates because quantities related to the circle can be expressed directly in terms of \(r\) and \(\theta\). For the disk of radius 3 centered at the origin, the transformation from Cartesian to polar coordinates is particularly helpful. We use the transformations \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\) to express \(x^2 + y^2 = r^2\). This greatly simplifies setting up and evaluating our integral.
Jacobian Determinant
In coordinate transformations, the Jacobian determinant plays a crucial role. It adjusts the "size" of infinitesimally small elements when switching coordinate systems, ensuring that integrals over transformed regions remain accurate.

When transforming from Cartesian to polar coordinates, the Jacobian determinant is \(r\). This adjustment term accounts for the fact that an infinitesimal area \(dA\) in Cartesian coordinates \(dA = dx \, dy\) relates to a slightly "expanded" area in polar coordinates \(dA = r \, dr \, d\theta\).

The inclusion of the Jacobian in the integral acknowledges that the grid "expands" as \(r\) increases, mimicking the increasing circumference of circles as they grow outward from the origin. Thus, it is essential to multiply the integrand by the Jacobian determinant \(r\) in the integral, transforming it to \(\iint_R e^{-r^2} r \, dr \, d\theta\). This ensures the geometric characteristics of the region are respected in the integral calculation.

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Most popular questions from this chapter

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 \(x=3 u\) \(y=2 v, z=w\)

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