Area of Circle

\[\begin{aligned} x^2+y^2&=r^2\\ y^2&=r^2-x^2\\ y&=\pm\sqrt{r^2-x^2} \end{aligned}\]

A=∫−rrr2−x2dx+∣∫−rr−r2−x2dx∣=∫−rrr2−x2dx+∣−∫−rrr2−x2dx∣=∫−rrr2−x2dx+∫−rrr2−x2dx=2∫−rrr2−x2dx, let x=rsin⁡θ⇒dx=rcos⁡θdθ=2∫−π2π2r2−(rsin⁡θ)2rcos⁡θdθ=2∫−π2π2r2−r2sin⁡2θrcos⁡θdθ=2∫−π2π2r2(1−sin⁡2θ)rcos⁡θdθ=2∫−π2π2r2cos⁡2θrcos⁡θdθ=2∫−π2π2rcos⁡θrcos⁡θdθ=2∫−π2π2r2cos⁡2θdθ=2r2∫−π2π2(12+12cos⁡2θ)dθ=2r2[12θ+14sin⁡2θ]−π2π2=2r2(12(π2)+14sin⁡(2(π2))−(12(−π2)+14sin⁡(2(−π2))))=2r2(π4+14sin⁡π+π4+14sin⁡π)=2r2(π2)A=πr2\begin{aligned} A&=\int_{-r}^{r}\sqrt{r^2-x^2}dx+\left|\int_{-r}^{r}-\sqrt{r^2-x^2}dx\right|\\ &=\int_{-r}^{r}\sqrt{r^2-x^2}dx+\left|-\int_{-r}^{r}\sqrt{r^2-x^2}dx\right|\\ &=\int_{-r}^{r}\sqrt{r^2-x^2}dx+\int_{-r}^{r}\sqrt{r^2-x^2}dx\\ &=2\int_{-r}^{r}\sqrt{r^2-x^2}dx\text{, let }x=r\sin\theta\Rightarrow dx=r\cos\theta d\theta\\ &=2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{r^2-(r\sin\theta)^2}r\cos\theta d\theta\\ &=2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{r^2-r^2\sin^2\theta}r\cos\theta d\theta\\ &=2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{r^2\left(1-\sin^2\theta\right)}r\cos\theta d\theta\\ &=2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{r^2}\sqrt{\cos^2\theta}r\cos\theta d\theta\\ &=2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}r\cos\theta r\cos\theta d\theta\\ &=2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}r^2\cos^2\theta d\theta\\ &=2r^2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac12+\frac12\cos2\theta\right)d\theta\\ &=2r^2\left[\frac12\theta+\frac14\sin2\theta\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\\ &=2r^2\left(\frac12\left(\frac{\pi}{2}\right)+\frac14\sin\left(2\left(\frac{\pi}{2}\right)\right)-\left(\frac12\left(-\frac{\pi}{2}\right)+\frac14\sin\left(2\left(-\frac{\pi}{2}\right)\right)\right)\right)\\ &=2r^2\left(\frac{\pi}{4}+\frac14\sin\pi+\frac{\pi}{4}+\frac14\sin\pi\right)\\ &=2r^2\left(\frac{\pi}{2}\right)\\ A&=\boxed{\pi r^2} \end{aligned}

#Calculus

Note by Gandoff Tan
1 year, 9 months ago

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Any proof for GCSE students, @Gandoff Tan?

A Former Brilliant Member - 1 year ago
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