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I got stuck solving this definite integral using trigonometric substitution

$$F(x)=\int_{-\sqrt{r^{2}-x^{2}}}^{\sqrt{r^{2}-x^{2}}}\sqrt{r^{2}-(x^{2}+y^{2})}dy$$ where $-r\leqslant x\leqslant r$

I let

$$y=r\sin{\theta}$$ $$dy=r\cos{\theta} d\theta$$ $$-\frac{\pi}{2}\leqslant \theta\leqslant \frac{\pi}{2}$$

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{r^{2}-(x^{2}+(r\sin{\theta})^{2})} r\cos{\theta} d\theta $$ $$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{r^{2}-x^{2}-(r^{2}(1-\cos^{2}{\theta}))} r\cos{\theta} d\theta $$ $$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{r^{2}\cos^{2}{\theta}-x^{2}} r\cos{\theta} d\theta $$

and then I got lost. Am I not seeing something?

kauselis3000
  • 263

2 Answers2

2

You need to think of the integrand as $\sqrt{(r^{2}-x^{2})-y^{2}}$ and substitute $y=\sqrt{r^2-x^2}\sin \theta$; then the substitution gives $$(r^2-x^2)\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^{2}\theta \, d\theta. $$

mcd
  • 3,448
1

If trigonometric substitution is optional, you may instead try an Euler substitution. Let

$$\begin{align*} t=\frac{\sqrt{r^2-x^2-y^2}-\sqrt{r^2-x^2}}y \implies y&=-2\sqrt{r^2-x^2}\frac{t}{1+t^2} \\ \implies dy &= -2\sqrt{r^2-x^2} \frac{1-t^2}{(1+t^2)^2} \, dt \end{align*}$$

so that

$$\begin{align*} F(x) &= \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \sqrt{r^2-x^2-y^2} \, dy \\ &= 2\sqrt{r^2-x^2} \int_{-1}^1 \sqrt{r^2-x^2-\frac{4(r^2-x^2)t^2}{(1+t^2)^2}} \, \frac{1-t^2}{(1+t^2)^2} \, dt \\ &= 2(r^2-x^2) \int_{-1}^1 \frac{(1-t^2)^2}{(1+t^2)^3} \, dt \end{align*}$$

user170231
  • 19,334