Solution assignment 05 Area and volume

Assignment 5

Calculate the area which is enclosed by the circle $x^2+y^2=R^2$ , that is a circle with center $(0,0)$ and radius $R$ .

Solution

The area of the circle can be calculated by first calculating the area of the circle in the first quadrant and multiply it by $4$ . The function in the first quadrant is:

$y=\sqrt{R^2-x^2}$

and thus we have to calculate:

$\displaystyle\int_{0}^{1}\sqrt{R^2-x^2}dx$

First we apply the substitution method:

$x=R\sin(\phi)$ , with $\phi$ ranging from $\phi=0$ to $\phi=\displaystyle\frac{\pi}{2}$

which results in:

$\displaystyle\int_{0}^{\frac{\pi}{2}}\sqrt{R^2-R^2\sin^2{(\phi)}}dR\sin{(\phi)}=R^2\displaystyle\int_{0}^{\frac{\pi}{2}}\cos^2{(\phi)}d\phi$

Now we use a formula from trigonometry:

$\cos(2\phi)=2\cos^2(\phi)-1$

and thus:

$\cos^2(\phi)=\displaystyle\frac{\cos(2\phi)+1}{2}=\displaystyle\frac{1}{2}\cos(2\phi)+\displaystyle\frac{1}{2}$

Then the integral becomes:

$R^2\displaystyle\int_{0}^{\frac{\pi}{2}}[\frac{1}{2}\cos(2\phi)+\frac{1}{2}]d\phi=R^2[\frac{1}{4}\sin(2\phi)+\frac{1}{2}\phi]_{0}^{\frac{\pi}{2}}=\frac{1}{4}{\pi}R^2$

The area of the whole circle is thus $4\cdot\displaystyle\frac{1}{4}{\pi}R^2={\pi}R^2$ . This is the result we already knew.

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Solution assignment 05 Area and volume

Assignment 5

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