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

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Tayeeba

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I don’t understand why this is symmetric.


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

. . . . .sin⁡(θ) = 12\displaystyle \sin(\theta)\, =\, \dfrac{1}{2}

...so:

. . . . .θ = π6, 5π6\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:

. . . . .A = 12 ∫π/65π/6 (3 sin⁡(θ))2 dθ − 12 ∫π/65π/6 (1 + sin⁡(θ))2 dθ\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 θ = π2,\displaystyle \theta\, =\, \frac{\pi}{2},, we can write:

. . . . .A = 2 [ 12 ∫π/6π/2 9 sin⁡2(θ) dθ − 12 ∫π/6π/2 (1 + 2 sin⁡(θ) + sin⁡2(θ)) dθ ]\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]

. . . . . . . . . .= ∫π/6π/2 (8 sin⁡2(θ) − 1 − 2 sin⁡(θ)) dθ\displaystyle \displaystyle =\, \int_{\pi/6}^{\pi/2}\, \left(8\, \sin^2(\theta)\, -\, 1\, -\, 2\, \sin(\theta)\right)\, d\theta

. . . . . . . . . .= ∫π/6π/2 (3 − 4 cos⁡(2 θ) − 2 sin⁡(θ)) dθ\displaystyle \displaystyle =\, \int_{\pi/6}^{\pi/2}\, \big(3\, -\, 4\, \cos(2\, \theta)\, -\, 2\, \sin(\theta)\big)\, d\theta

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


Any explanation please?
 

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Tayeeba said:
I don’t understand why this is symmetric.


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

. . . . .sin⁡(θ) = 12\displaystyle \sin(\theta)\, =\, \dfrac{1}{2}

...so:

. . . . .θ = π6, 5π6\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:

. . . . .A = 12 ∫π/65π/6 (3 sin⁡(θ))2 dθ − 12 ∫π/65π/6 (1 + sin⁡(θ))2 dθ\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 θ = π2,\displaystyle \theta\, =\, \frac{\pi}{2},, we can write:

. . . . .A = 2 [ 12 ∫π/6π/2 9 sin⁡2(θ) dθ − 12 ∫π/6π/2 (1 + 2 sin⁡(θ) + sin⁡2(θ)) dθ ]\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]

. . . . . . . . . .= ∫π/6π/2 (8 sin⁡2(θ) − 1 − 2 sin⁡(θ)) dθ\displaystyle \displaystyle =\, \int_{\pi/6}^{\pi/2}\, \left(8\, \sin^2(\theta)\, -\, 1\, -\, 2\, \sin(\theta)\right)\, d\theta

. . . . . . . . . .= ∫π/6π/2 (3 − 4 cos⁡(2 θ) − 2 sin⁡(θ)) dθ\displaystyle \displaystyle =\, \int_{\pi/6}^{\pi/2}\, \big(3\, -\, 4\, \cos(2\, \theta)\, -\, 2\, \sin(\theta)\big)\, d\theta

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


Any explanation please?
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I think you are asking why, as stated, the region between the cardioid r=1+sin(θ)\displaystyle r = 1 + sin(\theta) and the circle r=3sin(θ)\displaystyle r = 3 sin(\theta) is symmetrical about the vertical axis θ=π2\displaystyle \theta = \frac{\pi}{2}.

That is because the sine function is symmetrical about θ=π2\displaystyle \theta = \frac{\pi}{2}. That is, sin(π2+ϕ)=sin(π2−ϕ)\displaystyle sin(\frac{\pi}{2} + \phi) = sin(\frac{\pi}{2} - \phi). Think of the graph of y=sin(x)\displaystyle y = sin(x), which is symmetric about the line x=π2\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.
 
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Dr.Peterson said:
I think you are asking why, as stated, the region between the cardioid r=1+sin(θ)\displaystyle r = 1 + sin(\theta) and the circle r=3sin(θ)\displaystyle r = 3 sin(\theta) is symmetrical about the vertical axis θ=π2\displaystyle \theta = \frac{\pi}{2}.

That is because the sine function is symmetrical about θ=π2\displaystyle \theta = \frac{\pi}{2}. That is, sin(π2+ϕ)=sin(π2−ϕ)\displaystyle sin(\frac{\pi}{2} + \phi) = sin(\frac{\pi}{2} - \phi). Think of the graph of y=sin(x)\displaystyle y = sin(x), which is symmetric about the line x=π2\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.
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Thanks a lot
 
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