Convert to Polar Coordinates

Oct 13, 2025 by recoveryshake.com 35 views

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Introduction

In this article, we will evaluate the double integral $\iint_R(3x-y) \, dA$ , where $R$ is the region in the first quadrant enclosed by the circle $x^2 + y^2 = 9$ and the lines $x = 0$ and $y = x$ . To solve this problem, we will convert the given integral to polar coordinates, which will simplify the calculation and make it easier to evaluate.

Convert to Polar Coordinates

The region $R$ is enclosed by the circle $x^2 + y^2 = 9$ , which can be written in polar coordinates as $r^2 = 9$ , or $r = 3$ . The lines $x = 0$ and $y = x$ can be written in polar coordinates as $\theta = \frac{\pi}{2}$ and $r \cos \theta = r \sin \theta$ , respectively.

Express the Integral in Polar Coordinates

To express the integral in polar coordinates, we need to convert the function $3x-y$ to polar coordinates. We can do this by using the following transformations:

$x = r \cos \theta$ $y = r \sin \theta$

Substituting these transformations into the function $3x-y$ , we get:

$3x-y = 3r \cos \theta - r \sin \theta$

Evaluate the Integral

Now that we have expressed the integral in polar coordinates, we can evaluate it. The integral is given by:

$\iint_R(3x-y) \, dA = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{0}^{3} (3r \cos \theta - r \sin \theta) \, r \, dr \, d\theta$

Simplify the Integral

To simplify the integral, we can evaluate the inner integral first. We have:

$\int_{0}^{3} (3r \cos \theta - r \sin \theta) \, r \, dr = \int_{0}^{3} (3r^2 \cos \theta - r^2 \sin \theta) \, dr$

Evaluating the integral, we get:

$\int_{0}^{3} (3r^2 \cos \theta - r^2 \sin \theta) \, dr = \left[ r^3 \cos \theta - \frac{r^3}{3} \sin \theta \right]_{0}^{3}$

Evaluate the Outer Integral

Now that we have evaluated the inner integral, we can evaluate the outer integral. We have:

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left[ 3^3 \cos \theta - \frac{3^3}{3} \sin \theta \right] \, d\theta$

Evaluating the integral, we get:

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left[ 27 \cos \theta - 9 \sin \theta \right] \, d\theta = \left[ 27 \sin \theta + 9 \cos \theta \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$

Simplify the Result

Evaluating the expression, we get:

$\left[ 27 \sin \theta + 9 \cos \theta \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = (27 \sin \frac{\pi}{2} + 9 \cos \frac{\pi}{2}) - (27 \sin \frac{\pi}{4} + 9 \cos \frac{\pi}{4})$

Simplifying the expression, we get:

$(27 \sin \frac{\pi}{2} + 9 \cos \frac{\pi}{2}) - (27 \sin \frac{\pi}{4} + 9 \cos \frac{\pi}{4}) = (27 \cdot 1 + 9 \cdot 0) - (27 \cdot \frac{\sqrt{2}}{2} + 9 \cdot \frac{\sqrt{2}}{2})$

Simplifying the expression, we get:

$(27 \cdot 1 + 9 \cdot 0) - (27 \cdot \frac{\sqrt{2}}{2} + 9 \cdot \frac{\sqrt{2}}{2}) = 27 - \frac{27 \sqrt{2}}{2} - \frac{9 \sqrt{2}}{2}$

Simplifying the expression, we get:

$27 - \frac{27 \sqrt{2}}{2} - \frac{9 \sqrt{2}}{2} = 27 - \frac{36 \sqrt{2}}{2}$

Simplifying the expression, we get:

$27 - \frac{36 \sqrt{2}}{2} = 27 - 18 \sqrt{2}$

The final answer is $\boxed{27 - 18 \sqrt{2}}$ .

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Introduction

In the previous article, we evaluated the double integral $\iint_R(3x-y) \, dA$ , where $R$ is the region in the first quadrant enclosed by the circle $x^2 + y^2 = 9$ and the lines $x = 0$ and $y = x$ . We converted the integral to polar coordinates, which simplified the calculation and made it easier to evaluate. In this article, we will answer some common questions related to the double integral in polar coordinates.

Q: What is the region of integration in polar coordinates?

A: The region of integration in polar coordinates is given by $R = \{(r, \theta) | 0 \leq r \leq 3, \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}\}$ .

Q: How do I convert the function $3x-y$ to polar coordinates?

A: To convert the function $3x-y$ to polar coordinates, we use the following transformations:

$x = r \cos \theta$ $y = r \sin \theta$

Substituting these transformations into the function $3x-y$ , we get:

$3x-y = 3r \cos \theta - r \sin \theta$

Q: How do I evaluate the double integral in polar coordinates?

A: To evaluate the double integral in polar coordinates, we need to integrate the function with respect to $r$ and $\theta$ . We can do this by using the following formula:

$\iint_R f(x, y) \, dA = \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} f(r \cos \theta, r \sin \theta) \, r \, dr \, d\theta$

Q: What is the difference between polar coordinates and Cartesian coordinates?

A: Polar coordinates are a way of representing points in a plane using the distance from the origin and the angle from the positive x-axis. Cartesian coordinates, on the other hand, are a way of representing points in a plane using the x and y coordinates.

Q: When should I use polar coordinates instead of Cartesian coordinates?

A: You should use polar coordinates instead of Cartesian coordinates when the region of integration is a circle or an annulus, or when the function is more easily expressed in polar coordinates.

Q: How do I convert the double integral to polar coordinates?

A: To convert the double integral to polar coordinates, you need to convert the function and the region of integration to polar coordinates. You can do this by using the following transformations:

$x = r \cos \theta$ $y = r \sin \theta$

Q: What is the Jacobian in polar coordinates?

A: The Jacobian in polar coordinates is given by $r$ . This is because the area element in polar coordinates is given by $dA = r \, dr \, d\theta$ .

Q: How do I evaluate the Jacobian in polar coordinates?

A: To evaluate the Jacobian in polar coordinates, you need to find the partial derivatives of the function with respect to $r$ and $\theta$ . You can do this by using the following formula:

$\frac{\partial (f(r, \theta))}{\partial r} = \frac{\partial f}{\partial r}$ $\frac{\partial (f(r, \theta))}{\partial \theta} = \frac{\partial f}{\partial \theta}$

Q: What is the relationship between the Jacobian and the area element in polar coordinates?

A: The Jacobian is related to the area element in polar coordinates by the following formula:

$dA = r \, dr \, d\theta$

This means that the Jacobian is equal to the area element divided by $r$ .

Q: How do I use the Jacobian to evaluate the double integral in polar coordinates?

A: To use the Jacobian to evaluate the double integral in polar coordinates, you need to multiply the function by the Jacobian and then integrate with respect to $r$ and $\theta$ . You can do this by using the following formula:

$\iint_R f(x, y) \, dA = \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} f(r \cos \theta, r \sin \theta) \, r \, dr \, d\theta$

The final answer is $\boxed{27 - 18 \sqrt{2}}$ .