Calculus area prob: When finding the area, how is this symmetric?

[Tayeeba]

I don’t understand why this is symmetric.

---

EXAMPLE 2. .Find the area of the region that lies within the circle \(\displaystyle r\, =\, 3\, \sin(\theta)\) and outside the cardioid \(\displaystyle r\, =\, 1\, +\, \sin(\theta).\)

SOLUTION. .The cardioid (see Example 10.3.7) and the circle are sketched in Figure 5, and the desired region is shaded. The values of a and b in Formula 4 are determined by find the points of intersection of the two curves. They intersect when \(\displaystyle 3\, \sin(\theta)\, =\, 1\, +\, \sin(\theta),\) which gives:

. . . . .\(\displaystyle \sin(\theta)\, =\, \dfrac{1}{2}\)

...so:

. . . . .\(\displaystyle \theta\, =\, \dfrac{\pi}{6},\, \dfrac{5\pi}{6}\)

The desired area can be found by subtracting the area inside the cardioid between these intersection points from the area inside the circle between the same intersection points. Thus:

. . . . .\(\displaystyle \displaystyle A\, =\, \frac{1}{2}\, \int_{\pi/6}^{5\pi/6}\, \big(3\, \sin(\theta)\big)^2\, d\theta\, -\, \frac{1}{2}\, \int_{\pi/6}^{5\pi/6}\, \big(1\, +\, \sin(\theta)\big)^2\, d\theta\)

Since the region is symmetric about the vertical axis \(\displaystyle \theta\, =\, \frac{\pi}{2},\), we can write:

. . . . .\(\displaystyle \displaystyle A\, =\, 2\, \bigg[\, \frac{1}{2}\, \int_{\pi/6}^{\pi/2}\, 9\, \sin^2(\theta)\, d\theta\, -\, \frac{1}{2}\, \int_{\pi/6}^{\pi/2}\, \left(1\, +\, 2\, \sin(\theta)\, +\, \sin^2(\theta)\right)\, d\theta \, \bigg]\)

. . . . . . . . . .\(\displaystyle \displaystyle =\, \int_{\pi/6}^{\pi/2}\, \left(8\, \sin^2(\theta)\, -\, 1\, -\, 2\, \sin(\theta)\right)\, d\theta\)

. . . . . . . . . .\(\displaystyle \displaystyle =\, \int_{\pi/6}^{\pi/2}\, \big(3\, -\, 4\, \cos(2\, \theta)\, -\, 2\, \sin(\theta)\big)\, d\theta\)

. . . . . . . . . .\(\displaystyle =\, 3\,\theta\, -\, 2\, \sin(2\, \theta)\, +\, 2\, \cos(\theta)\bigg]_{\pi/6}^{\pi/2}\, =\, \pi\)

---

Any explanation please?

 

[Dr.Peterson]

I don’t understand why this is symmetric.

---

EXAMPLE 2. .Find the area of the region that lies within the circle \(\displaystyle r\, =\, 3\, \sin(\theta)\) and outside the cardioid \(\displaystyle r\, =\, 1\, +\, \sin(\theta).\)

SOLUTION. .The cardioid (see Example 10.3.7) and the circle are sketched in Figure 5, and the desired region is shaded. The values of a and b in Formula 4 are determined by find the points of intersection of the two curves. They intersect when \(\displaystyle 3\, \sin(\theta)\, =\, 1\, +\, \sin(\theta),\) which gives:

. . . . .\(\displaystyle \sin(\theta)\, =\, \dfrac{1}{2}\)

...so:

. . . . .\(\displaystyle \theta\, =\, \dfrac{\pi}{6},\, \dfrac{5\pi}{6}\)

The desired area can be found by subtracting the area inside the cardioid between these intersection points from the area inside the circle between the same intersection points. Thus:

. . . . .\(\displaystyle \displaystyle A\, =\, \frac{1}{2}\, \int_{\pi/6}^{5\pi/6}\, \big(3\, \sin(\theta)\big)^2\, d\theta\, -\, \frac{1}{2}\, \int_{\pi/6}^{5\pi/6}\, \big(1\, +\, \sin(\theta)\big)^2\, d\theta\)

Since the region is symmetric about the vertical axis \(\displaystyle \theta\, =\, \frac{\pi}{2},\), we can write:

. . . . .\(\displaystyle \displaystyle A\, =\, 2\, \bigg[\, \frac{1}{2}\, \int_{\pi/6}^{\pi/2}\, 9\, \sin^2(\theta)\, d\theta\, -\, \frac{1}{2}\, \int_{\pi/6}^{\pi/2}\, \left(1\, +\, 2\, \sin(\theta)\, +\, \sin^2(\theta)\right)\, d\theta \, \bigg]\)

. . . . . . . . . .\(\displaystyle \displaystyle =\, \int_{\pi/6}^{\pi/2}\, \left(8\, \sin^2(\theta)\, -\, 1\, -\, 2\, \sin(\theta)\right)\, d\theta\)

. . . . . . . . . .\(\displaystyle \displaystyle =\, \int_{\pi/6}^{\pi/2}\, \big(3\, -\, 4\, \cos(2\, \theta)\, -\, 2\, \sin(\theta)\big)\, d\theta\)

. . . . . . . . . .\(\displaystyle =\, 3\,\theta\, -\, 2\, \sin(2\, \theta)\, +\, 2\, \cos(\theta)\bigg]_{\pi/6}^{\pi/2}\, =\, \pi\)

---

Any explanation please?

I think you are asking why, as stated, the region between the cardioid \(\displaystyle r = 1 + sin(\theta)\) and the circle \(\displaystyle r = 3 sin(\theta)\) is symmetrical about the vertical axis \(\displaystyle \theta = \frac{\pi}{2}\).

That is because the sine function is symmetrical about \(\displaystyle \theta = \frac{\pi}{2}\). That is, \(\displaystyle sin(\frac{\pi}{2} + \phi) = sin(\frac{\pi}{2} - \phi)\). Think of the graph of \(\displaystyle y = sin(x)\), which is symmetric about the line \(\displaystyle x = \frac{\pi}{2}\).

Or you can just look at the figure. It is implied that they have previously discussed the cardioid (Example 10.3.7), so there might be more information there.

 

[Tayeeba]

I think you are asking why, as stated, the region between the cardioid \(\displaystyle r = 1 + sin(\theta)\) and the circle \(\displaystyle r = 3 sin(\theta)\) is symmetrical about the vertical axis \(\displaystyle \theta = \frac{\pi}{2}\).

That is because the sine function is symmetrical about \(\displaystyle \theta = \frac{\pi}{2}\). That is, \(\displaystyle sin(\frac{\pi}{2} + \phi) = sin(\frac{\pi}{2} - \phi)\). Think of the graph of \(\displaystyle y = sin(x)\), which is symmetric about the line \(\displaystyle x = \frac{\pi}{2}\).

Or you can just look at the figure. It is implied that they have previously discussed the cardioid (Example 10.3.7), so there might be more information there.

Thanks a lot