### Math Notes

Subjects

#### Integral Calculus Solutions

##### Topics || Problems

Find the area enclosed by r = 2a Cos θ

Solution:

Differential Area

$$dA = \frac{1}{2}{r^2}d\theta$$

$$dA = \frac{1}{2}{\left( {2aCos\theta } \right)^2}d\theta$$

Limits:

Let $$r = 0$$

$$\cos \theta = 0$$

$$\theta = \frac{\pi }{2},\frac{{3\pi }}{2}$$

Since it's symmetrical along the x axis. The limits can also be from 0 to $$\frac{\pi }{2}$$ and multiply the area twice.

Thus:

$$A = \frac{1}{2}\int\limits_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {4{a^2}{{\cos }^2}\theta d\theta }$$

$$A = 2{a^2}\int\limits_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\cos }^2}\theta d\theta }$$

$${\cos ^2}\theta d\theta = \frac{1}{2}\left( {1 + \cos 2\theta } \right)$$

$$A = 2{a^2}\int\limits_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {\frac{1}{2}\left( {1 + \cos 2\theta } \right)d\theta }$$

$$A = {a^2}\left[ {\theta + \frac{1}{2}\sin 2\theta } \right]_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}}$$

$$A = {a^2}\left[ {\left( {\frac{\pi }{2} + 0} \right) - \left( {\frac{{ - \pi }}{2} + 0} \right)} \right]$$

$$A = \pi {a^2}$$ square unitsAnswer