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Question Number 33823 by 33 last updated on 25/Apr/18

$$solve:$$ $$I=\underset{0}{\stackrel{\pi }{\int }}\frac{(r-Rcos\theta )sin\theta }{{(R^2+r^2-2Rrcos\theta )}^{3/2}}d\theta$$ $$forr<R$$ $$andr>Rrespectively.$$

Answered by MJS last updated on 26/Apr/18

$$\int u^{\prime }v=uv-\int {uv}^{\prime }$$ $$u^{\prime }=\frac{\sin \theta }{{(r^2+R^2-2rR\cos \theta )}^{\frac{3}{2}}};v=r-R\cos \theta$$ $$u=-\frac{1}{rR{(r^2+R^2-2rR\cos \theta )}^{\frac{1}{2}}};v^{\prime }=R\sin \theta$$ $$\left[\begin{array}{c}\int \frac{\sin \theta }{{(r^2+R^2-2rR\cos \theta )}^{\frac{3}{2}}}d\theta =\\ [t=r^2+R^2-2rR\cos \theta \to dx=\frac{dt}{2rR\sin \theta }]\\ =\frac{1}{2rR}\int \frac{1}{t^{\frac{3}{2}}}dt=-\frac{1}{{rRt}^{\frac{1}{2}}}\end{array}\right]$$ $$uv=-\frac{r-R\cos \theta }{rR{(r^2+R^2-2rR\cos \theta )}^{\frac{1}{2}}}$$ $$-\int {uv}^{\prime }=\int \frac{\sin \theta }{r{(r^2+R^2-2rR\cos \theta )}^{\frac{1}{2}}}=$$ $$\left[\mathrm{with}\mathrm{the}\mathrm{same}\mathrm{substitution}\mathrm{as}\mathrm{above}\right]$$ $$=\frac{{(r^2+R^2-2rR\cos \theta )}^{\frac{1}{2}}}{r^{2}R}$$ $$\int \frac{(r-R\cos \theta )\sin \theta }{{(r^2+R^2-2rR\cos \theta )}^{\frac{3}{2}}}=$$ $$=-\frac{r-R\cos \theta }{rR{(r^2+R^2-2rR\cos \theta )}^{\frac{1}{2}}}+\frac{{(r^2+R^2-2rR\cos \theta )}^{\frac{1}{2}}}{r^{2}R}+C=$$ $$=\frac{R-r\cos \theta }{r^2\sqrt{r^2+R^2-2rR\cos \theta }}+C=F\left(\theta \right)$$ $$F\left(\pi \right)-F\left(0\right)=\frac{R+r}{r^2\mid R+r\mid }-\frac{R-r}{r^2\mid R-r\mid }$$ $$r,R>0$$ $$r<R\Rightarrow F\left(\pi \right)-F\left(0\right)=0$$ $$r>R\Rightarrow F\left(\pi \right)-F\left(0\right)=\frac{2}{r^2}$$

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