Using integral to find the area under a portion of a semicircle

Question

I've done quite a bit of research, but I always fail right before the end. I know the integral of a circle, I know the substitution $$\theta = \arcsin\left(\frac x r\right)$$ However, I never get the right results.

My question is:

How do I find the area under a semicircle from $-r$ to $x_1; x_1 \in[-r;+r]$?

Also if I want to find the are under a semicircle offset in $$y= f(x) = \sqrt{r^2 - x^2} - d$$

I would simply have to use the above formular and subtract $(x + r) * d$?

Answer 1

$$A=\int_{-r}^{x_1}\sqrt{r^2-x^2}dx$$ Let $x = r\sin\theta$
$dx=r\cos\theta$

$$\begin{align} A&=\int_{-\pi/2}^{\arcsin{(x_1/r)}}(r\cos^2\theta )d\theta\\ &=r\int_{-\pi/2}^{\arcsin{(x_1/r)}}{\cos2\theta+1\over2}d\theta\\ &=\frac r2\left[\frac12\sin2\theta+\theta\right]_{-\pi/2}^{\arcsin(x_1/r)}\\ &=\frac r2\left[\sin\theta\cos\theta+\theta\right]_{-\pi/2}^{\arcsin(x_1/r)}\\ &=\frac r2\left({x_1\sqrt{r^2-x_1^2}\over r^2} +\arcsin(x_1/r)+\frac\pi2 \right) \end{align}$$

Not sure what you mean by second part but just place the center on origin and calculate normally. Just be sure to change $r_1$ and $x$ accordingly

Answer 2

You can consider $$\cases{x=r\cos\theta\\y=r\sin\theta}$$($x$ and $y$ are interchangable for circles at the standard position)

So $$dx=-r\sin\theta \ d\theta$$

So your area would be

\begin{align} \int_{-r}^{x_1}(y-0) \ dx &=\int_{\pi}^{\theta_1}(r\sin\theta)(-r\sin\theta) \ d\theta; &x_1=r\cos\theta_1\\ &=r^2\int_{\theta_1}^\pi\sin^2\theta \ d\theta\\ &=\frac12r^2\int_{\theta_1}^\pi(1-\cos 2\theta) \ d\theta\\ &=\frac12r^2\left[\theta-\frac12\sin2\theta\right]_{\theta_1}^{\pi}\\ &=\frac12r^2\left[\frac\pi2-\left(\theta_1-\frac12\sin2\theta_1\right)\right]\\ &=r^2\left(\frac{\pi}{4}-\frac{1}2\theta_1+\frac14\sin2\theta_1\right) \end{align}

Answer 3

Hint:

Note that the searched area can be calculated without integration.

If $P=\left(x_1,\sqrt{r^2-x_1^2}\right)$ is the point on the circle, $O$ is the center of the circle, $A$ the projection of $P$ on the $x$ axis and $B=(r,0)$, than the area is:

Area$=($Area of semicircle$)-($Area of sector OPB$) +($Area triangle OPA$)$.

Or, if you consider the point $C=(0,r)$ we have:

Area$=($Area of a quart of circle$)+($Area of sector OCP$) +($Area triangle OPA$)$.

That can be done with elementary geometry.

The same method, with a bit of care, solve also the case of a shifted semicircle.

Answer 4

Integrate the area by parts directly

\begin{align} Area &=\int_{-r}^{x_1}\sqrt{r^2-x^2}dx =\int_{-r}^{x_1}\frac{\sqrt{r^2-x^2}}{2x}d(x^2)\\ & = \frac12 x \sqrt{r^2-x^2}\bigg|_{-r}^{x_1} + \frac12\int_{-r}^{x_1}\frac{r^2}{\sqrt{r^2-x^2}}dx\\ &= \frac12 x_1\sqrt{r^2-x_1^2}+ \frac12r^2 \sin^{-1}\frac xr \bigg|_{-r}^{x_1}\\ &= \frac12 x_1\sqrt{r^2-x_1^2}+ \frac12r^2 (\sin^{-1}\frac {x_1}r +\frac\pi2) \end{align}