Polar Coordinates and the Double Integral

Oct 13, 2025 by thesweetdesigns.com 138 views

Introduction

In this article, we will evaluate the integral of a polar function over a specific region. The region $R$ is defined in polar coordinates as $0 \leq r \leq 8$ and $0 < \theta \leq \frac{\pi}{2}$ . The integral to be evaluated is $\iint_R 14\left(1-x^2-y^2\right) \, dA$ . We will use the properties of polar coordinates and the definition of the double integral to simplify the integral and find its value.

Polar Coordinates and the Double Integral

In polar coordinates, a point in the plane is represented by the ordered pair $(r, \theta)$ , where $r$ is the distance from the origin to the point and $\theta$ is the angle between the positive $x$ -axis and the line segment connecting the origin to the point. The region $R$ is defined as the set of all points $(r, \theta)$ such that $0 \leq r \leq 8$ and $0 < \theta \leq \frac{\pi}{2}$ .

The double integral of a function $f(x,y)$ over a region $R$ is defined as $\iint_R f(x,y) \, dA = \lim_{\Delta A \to 0} \sum_{i=1}^n f(x_i, y_i) \Delta A$ , where $\Delta A$ is the area of a small subregion of $R$ and the sum is taken over all subregions.

In polar coordinates, the area element $dA$ is given by $dA = r \, dr \, d\theta$ . Therefore, the double integral of a function $f(r, \theta)$ over a region $R$ is given by $\iint_R f(r, \theta) \, dA = \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} f(r, \theta) r \, dr \, d\theta$ .

Evaluating the Integral

The integral to be evaluated is $\iint_R 14\left(1-x^2-y^2\right) \, dA$ . We can rewrite this integral in polar coordinates as $\iint_R 14\left(1-r^2\right) \, dA$ . Using the definition of the double integral in polar coordinates, we have:

$\iint_R 14\left(1-r^2\right) \, dA = \int_{0}^{\frac{\pi}{2}} \int_{0}^{8} 14\left(1-r^2\right) r \, dr \, d\theta$

Simplifying the Integral

We can simplify the integral by evaluating the inner integral first. We have:

$\int_{0}^{8} 14\left(1-r^2\right) r \, dr = \int_{0}^{8} 14r - 14r^3 \, dr$

Using the power rule of integration, we have:

$\int_{0}^{8} 14r - 14r^3 \, dr = \left[7r^2 - 2r^4\right]_{0}^{8}$

Evaluating the expression at the limits of integration, we have:

$\left[7r^2 - 2r^4\right]_{0}^{8} = 7(8)^2 - 2(8)^4 = 448 - 2048 = -1600$

Evaluating the Outer Integral

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

$\int_{0}^{\frac{\pi}{2}} -1600 \, d\theta = \left[-1600\theta\right]_{0}^{\frac{\pi}{2}}$

Evaluating the expression at the limits of integration, we have:

$\left[-1600\theta\right]_{0}^{\frac{\pi}{2}} = -1600\left(\frac{\pi}{2}\right) + 0 = -800\pi$

Conclusion

In this article, we evaluated the integral of a polar function over a specific region. The region $R$ is defined in polar coordinates as $0 \leq r \leq 8$ and $0 < \theta \leq \frac{\pi}{2}$ . The integral to be evaluated is $\iint_R 14\left(1-x^2-y^2\right) \, dA$ . We used the properties of polar coordinates and the definition of the double integral to simplify the integral and find its value.

The final answer is $\boxed{-800\pi}$ .

Introduction

In our previous article, we evaluated the integral of a polar function over a specific region. The region $R$ is defined in polar coordinates as $0 \leq r \leq 8$ and $0 < \theta \leq \frac{\pi}{2}$ . The integral to be evaluated is $\iint_R 14\left(1-x^2-y^2\right) \, dA$ . In this article, we will answer some common questions related to the evaluation of the integral.

Q: What is the definition of the double integral in polar coordinates?

A: The double integral of a function $f(r, \theta)$ over a region $R$ in polar coordinates is given by $\iint_R f(r, \theta) \, dA = \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} f(r, \theta) r \, dr \, d\theta$ .

Q: How do I convert a function from Cartesian coordinates to polar coordinates?

A: To convert a function from Cartesian coordinates to polar coordinates, you need to use the following transformations:

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

Q: What is the area element $dA$ in polar coordinates?

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

Q: How do I evaluate the integral of a function over a region in polar coordinates?

A: To evaluate the integral of a function over a region in polar coordinates, you need to follow these steps:

Convert the function from Cartesian coordinates to polar coordinates.
Define the region $R$ in polar coordinates.
Evaluate the inner integral first.
Evaluate the outer integral.

Q: What is the final answer to the integral $\iint_R 14\left(1-x^2-y^2\right) \, dA$ ?

A: The final answer to the integral $\iint_R 14\left(1-x^2-y^2\right) \, dA$ is $\boxed{-800\pi}$ .

Q: Can I use the same method to evaluate the integral of a function over a region in Cartesian coordinates?

A: Yes, you can use the same method to evaluate the integral of a function over a region in Cartesian coordinates. However, you need to use the area element $dA = dx \, dy$ instead of $dA = r \, dr \, d\theta$ .

Q: What are some common mistakes to avoid when evaluating the integral of a function over a region in polar coordinates?

A: Some common mistakes to avoid when evaluating the integral of a function over a region in polar coordinates include:

Forgetting to convert the function from Cartesian coordinates to polar coordinates.
Forgetting to define the region $R$ in polar coordinates.
Evaluating the inner integral first without checking the limits of integration.
Evaluating the outer integral without checking the limits of integration.

Conclusion

In this article, we answered some common questions related to the evaluation of the integral of a polar function over a region. We hope that this article has been helpful in clarifying any doubts you may have had. If you have any further questions, please don't hesitate to ask.