# Double integral in polar coordination

In summary, the conversation discusses using polar coordinates for a double integral and the correct ranges for integration. The main issue is determining the correct ranges, with one person questioning and the other confirming the ranges. They also discuss the result being negative and how it should always be positive.

## The Attempt at a Solution

Please tell me if I am wrong. I suspect about the ranges. Are my range corrrect?

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Please tell me if I am wrong. I suspect about the ranges. Are my range corrrect?

If you have a double integral on the form I = $$\int \int_{R} f(x,y) dy dx$$
Then if you to integrate using polar coordinates then it changes to
I = $$\int \int_{R} f(x,y) dA$$ where

$$dA = r dr d\theta$$

trying adding this and see if you don't get the right result.

Dear Susanne217,

Thank you very much for your reply, but i have already added $$dA = r dr d\theta$$
to get the result. If you calculate it you will see that I had already added it! But my problem is the ranges. I suspect about the ranges.

The quantity you integrate (that is '1') is always positive.
An integral is basically a sum, so a sum of positive terms cannot turn into something negative.
The result of the book $$(1-\sqrt2 \cong -0.41)$$ is negative, so it is wrong.

$$=\int_0^{\pi/4} \int_{0}^{\frac{sen\theta}{cos^2\theta}} dr\: d\theta$$

$$=\int_0^{\pi/4} {\frac{sen\theta}{cos^2\theta}} \: d\theta$$

$$=\left[{\frac{1}{cos\theta}\right]_0^{\pi/4}$$

$$=\sqrt2-1$$

Last edited:
Quinzio said:
The quantity you integrate (that is '1') is always positive.
An integral is basically a sum, so a sum of positive terms cannot turn into something negative.
The result of the book $$(1-\sqrt2 \cong -0.41)$$ is negative, so it is wrong.

Thank you very much for the notification about "positive". But would you please tell me what your idea is about the ranges? especially about the ranges of "r". Are you sure that my ranges are correct?

Yes they are correct because in a polar-cartesian system:
the system
$$y = x tg\theta$$
$$y = x^2$$

gives $$x = tg\theta$$

$$r = \sqrt(x^2+y^2)= tg\theta \sqrt(tg^2\theta+1) = tg\theta / cos\theta = sen\theta / cos^2\theta$$

Dear Quinzio,

Thank you very much for your help.

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## 1. What is a double integral in polar coordination?

A double integral in polar coordination is a type of integral that is used to find the volume or area of a region in the polar coordinate system. It involves integrating over a two-dimensional shape that is expressed in terms of polar coordinates (r, θ).

## 2. How is a double integral in polar coordination different from a regular double integral?

A regular double integral involves integrating over a two-dimensional shape in the Cartesian coordinate system (x, y). In contrast, a double integral in polar coordination involves integrating over a two-dimensional shape in the polar coordinate system (r, θ).

## 3. What are the limits of integration for a double integral in polar coordination?

The limits of integration for a double integral in polar coordination depend on the shape of the region being integrated. The inner integral typically has limits of 0 to 2π, representing a full revolution around the origin. The outer integral has limits that correspond to the outer boundary of the region.

## 4. What is the formula for calculating a double integral in polar coordination?

The formula for calculating a double integral in polar coordination is ∫∫f(r, θ)rdrdθ, where f(r, θ) is the function being integrated and r and θ are the polar coordinates.

## 5. What are some real-world applications of double integrals in polar coordination?

Double integrals in polar coordination are commonly used in physics and engineering to calculate the mass, moment of inertia, and center of mass of objects with circular or cylindrical symmetry. They are also used in calculus to solve problems related to areas, volumes, and surface areas in polar coordinates.

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