Question

Lines y=2x/3 & y=5 cut the ring formed by circles (x-3)^2+(y-5)=64 and (x-3)^2+(y-5)^2=25 into four parts. Find the area of each of the four parts.

Answer 1

Lines $y = 2x / 3$ ; $y = 5$ cut the ring formed by circles $(x - 3)^\wedge 2 + (y - 5)^\wedge 2 = 64$ and $(x - 3)^\wedge 2 + (y - 5)^\wedge 2 = 25$ into four parts. Find the area of each of the four parts.

We have

$$S _ {I} + S _ {I I} = S _ {I I I} + S _ {I V} = \pi \left(8 ^ {2} - 5 ^ {2}\right) / 2 = \frac {3 9 \pi}{2}$$

Let's find points of intersection

A(-5,5)

E(11,5)

B(-2,5)

F(8,5)

$\mathrm{D}\left(\frac{3}{13}\left(19 - \sqrt{751}\right),\frac{2}{13}\left(19 - \sqrt{751}\right)\right)$

$\mathrm{J}\left(\frac{1}{2}\left(\frac{114}{13} +\frac{6\sqrt{751}}{13}\right),\frac{2}{13}\left(19 + \sqrt{751}\right)\right)$

$\mathrm{C}\left(\frac{3}{13}\left(19 - 2\sqrt{61}\right),\frac{2}{13}\left(19 - 2\sqrt{61}\right)\right)$

$\mathrm{G}\left(\frac{1}{2}\left(\frac{114}{13} +\frac{12\sqrt{61}}{13}\right),\frac{2}{13}\left(19 + 2\sqrt{61}\right)\right)$

$$y - 5 = \pm \sqrt {6 4 - (x - 3) ^ {2}}$$ $$y - 5 = \pm \sqrt {2 5 - (x - 3) ^ {2}}$$ $$\begin{array}{l} S_{IV} = \int_{-5}^{3} \sqrt{64 - (x - 3)^2} dx - \int_{-2}^{3} \left(5 - \left(-\sqrt{25 - (x - 3)^2}\right) - 5\right) dx - \\ - \int_{\frac{3}{13} \left(19 - 2\sqrt{61}\right)}^{3} \left(\frac{2x}{3} + \sqrt{64 - (x - 3)^2} - 5\right) dx \approx 19.1654 \\ \end{array}$$

Then

$$S_{III} \approx 42.0956$$

Similar

$$\begin{array}{l} S_{II} = \int_{\frac{1}{2} \left(\frac{114}{13}, \frac{12\sqrt{61}}{13}\right)}^{11} \sqrt{64 - (x - 3)^2} dx - \int_{\frac{1}{2} \left(\frac{114}{13}, \frac{12\sqrt{61}}{13}\right)}^{8} \left(\sqrt{25 - (x - 3)^2} + 5 - 5\right) dx - \\ - \int_{\frac{1}{2} \left(\frac{114}{13}, \frac{12\sqrt{61}}{13}\right)}^{1} \left(\sqrt{64 - (x - 3)^2} + 5 - \frac{2x}{3}\right) dx \approx 3.76674 \\ \end{array}$$ $$S_I = 57.4943$$

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