Double integral with polar coordinates

In summary: It looks like you had no space before \begin and left off the ## delimiters at start/end. I fixed it but you deleted it as I was fixing it...
  • #1
Amaelle
310
54
Homework Statement
look at the image
Relevant Equations
polar coordinates
Greetings!
I have the following integral
1630000459327.png

and here is the solution of the book (which I understand perfectly)
1630000520759.png

1630000549035.png


I have an altenative method I want to apply that does not seems to gives me the final resultMy method
1630002484185.png

which doesn't give me the final results!
where is my mistake?
thank you!
 
Physics news on Phys.org
  • #2
You can write \begin{align*}
I &= \int_0^{\pi/4} \int_0^{\sqrt{2+2\cos{2\theta}}} 3r^2 dr d\theta \\
&= \int_0^{\pi/4} \cos{\theta} \left( 2+2\cos{2\theta}\right)^{3/2} d\theta \\
&= 8 \int_0^{\pi/4} \cos^4{\theta} d\theta
\end{align*}then\begin{align*}
\int \cos^4{x} dx &=\int \dfrac{(e^{ix} + e^{-ix})^4}{16} dx \\
&= \int \left(\dfrac{e^{4ix} + e^{-4ix} + 4e^{2ix} + 4e^{-2ix} + 6}{16} \right) dx \\
&= \int \left( \dfrac{1}{8} \cos{4x} + \dfrac{1}{2} \cos{2x} + \dfrac{3}{8} \right) dx \\
&= \dfrac{1}{32} \sin{4x} + \dfrac{1}{4} \sin{2x} + \dfrac{3}{8} x \bigg{|}_{0}^{\pi/4} \\
&= \dfrac{1}{4} + \dfrac{3\pi}{32}
\end{align*}
 
Last edited:
  • #3
ergospherical said:
You can write
\begin{align*}
I &= \int_0^{\pi/4} \int_0^{\sqrt{2+2\cos{2\theta}}} 3r^2 dr d\theta \\
&= \int_0^{\pi/4} \cos{\theta} \left( 2+2\cos{2\theta}\right)^{3/2} d\theta \\
&= 8 \int_0^{\pi/4} \cos^4{\theta} d\theta
\end{align*}
then
\begin{align*}
\int \cos^4{x} dx &=\int \dfrac{(e^{ix} + e^{-ix})^4}{16} dx \\
&= \int \left(\dfrac{e^{4ix} + e^{-4ix} + 4e^{2ix} + 4e^{-2ix} + 6}{16} \right) dx \\
&= \int \left( \dfrac{1}{8} \cos{4x} + \dfrac{1}{2} \cos{2x} + \dfrac{3}{8} \right) dx \\
&= \dfrac{1}{32} \sin{4x} + \dfrac{1}{4} \sin{2x} + \dfrac{3}{8} x \bigg{|}_{0}^{\pi/4} \\
&= \dfrac{1}{4} + \dfrac{3\pi}{32}
\end{align*}
thank you but your code is not readable
 
  • #4
Amaelle said:
thank you but your code is not readable
Yeah I don't know why that's happening, the LaTeX is fine and works when I run it on Overleaf. :frown:
 
  • Sad
Likes Amaelle
  • #5
@ergospherical thank you very much!
indeed my approach was correct, I only messed up with the calculation when I was trying to integrate 8*cos^5 instead of 8*cos^4
thank you again!
 
  • #6
ergospherical said:
Yeah I don't know why that's happening, the LaTeX is fine and works when I run it on Overleaf
It looks like you had no space before \begin and left off the ## delimiters at start/end. I fixed it but you deleted it as I was fixing it...
 
  • Like
Likes SammyS

FAQ: Double integral with polar coordinates

What is a double integral with polar coordinates?

A double integral with polar coordinates is a mathematical tool used to calculate the area under a curved surface in a two-dimensional polar coordinate system. It involves integrating a function over a region in the polar coordinate plane.

What is the difference between a regular double integral and a double integral with polar coordinates?

The main difference between a regular double integral and a double integral with polar coordinates is the coordinate system used. In a regular double integral, the coordinates are expressed in terms of x and y, while in a double integral with polar coordinates, the coordinates are expressed in terms of r (radius) and θ (angle).

How do you convert a regular double integral into a double integral with polar coordinates?

To convert a regular double integral into a double integral with polar coordinates, the limits of integration must be changed from rectangular coordinates to polar coordinates. The function being integrated must also be expressed in terms of r and θ.

What are the advantages of using a double integral with polar coordinates?

One advantage of using a double integral with polar coordinates is that it can simplify the calculation of areas under curved surfaces, especially when the region has circular or symmetrical boundaries. It can also be useful in solving problems involving polar coordinates, such as calculating moments of inertia.

What are some real-life applications of double integrals with polar coordinates?

Double integrals with polar coordinates have many real-life applications, including calculating the electric field of a charged ring, finding the center of mass of a circular plate, and determining the volume of a cone or sphere. They are also used in engineering, physics, and other fields to solve problems involving polar coordinates.

Back
Top