How to show two identical integrals $\; 2\int_{-r}^{r}f(x)dx=r\int_{-r}^{r}\sqrt{1+(df/dx)^{2}}dx$

Question

With $$ \;f(x)=(r^2-x^2)^{1/2}$$

show that$$\; 2\int_{-r}^{r}f(x)dx=r\int_{-r}^{r}\sqrt{1+(df/dx)^{2}}dx$$

by using integration by parts. Then, conclude that the area of a circle of radius r is $\;\pi r^2$ if we define $\;\pi=\;$ circumference$\;\div\;$ diameter

I thought the only method I can use is via trig substitutions. But, the problem asks to solve it with integration-by-parts.

I need help to solve this problem. Thank you.

Answer 1

The equation claims the area of the centre-$O$ radius-$r$ circle is $r$ times the half/circumference due to its upper semicircle, i.e. $\pi r^2=r\cdot\tfrac12\cdot2\pi r$, which is trivial if you know the semicircle's area is $\tfrac12\pi r^2$. The challenge at hand is to prove this by IBP instead. First write the problem as$$\int_{-r}^r2\sqrt{r^2-x^2}dx=\int_{-r}^r\frac{r^2}{\sqrt{r^2-x^2}}dx.$$Now integrate$$2\sqrt{r^2-x^2}=\sqrt{r^2-x^2}-\frac{x^2}{\sqrt{r^2-x^2}}+\frac{r^2}{\sqrt{r^2-x^2}}=\left(x\sqrt{r^2-x^2}\right)^\prime+\frac{r^2}{\sqrt{r^2-x^2}}.$$

Answer 2

Integrate by parts as follows \begin{align} LHS= &\> 2\int_{-r}^{r}\sqrt{r^2-x^2} dx =\int_{-r}^{r}\frac{\sqrt{r^2-x^2}}xd(x^2)\\ = & \>x \sqrt{r^2-x^2}\bigg|_{-r}^{r} + \int_{-r}^{r}\frac{r^2}{\sqrt{r^2-x^2}}dx\\ =&\>0+\>r\int_{-r}^{r}\sqrt{1+\left(\frac{df }{dx}\right)^{2}}dx=RHS \end{align}

Answer 3

$$f(x)=(r^2-x^2)^{1/2}$$ $$f'(x)=-x(r^2-x^2)^{-1/2}$$ $$f'^2(x)=\frac{x^2}{r^2-x^2}$$ $$1+f'^2(x)=1+\frac{x^2}{r^2-x^2}=\frac{r^2-x^2+x^2}{r^2-x^2}=\frac{r^2}{r^2-x^2}$$ $$\sqrt{1+f'^2(x)}=r(r^2-x^2)^{-1/2}$$ so: $$r\int_{-r}^r\sqrt{1+f'^2(x)}dx=r^2\int_{-r}^r(r^2-x^2)^{-1/2}dx$$ and: $$2\int_{-r}^rf(x)dx=2\int_{-r}^r(r^2-x^2)^{1/2}$$ can u finish it from here?