Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 60987 by ajfour last updated on 28/May/19

Commented by ajfour last updated on 28/May/19

$$Findtheilluminatedareaof$$$$theinnercurvedsurfaceof$$$$shown,hollowopencylinder.$$$$[wherea>2R(\frac{H}{h}-1)]$$

Answered by mr W last updated on 28/May/19

Commented by mr W last updated on 30/May/19

$$\tan \alpha =\frac{R\sin \theta }{a+R(1+\cos \theta )}=\frac{\sin \theta }{\frac{a}{R}+1+\cos \theta }$$$${\theta }_0=\frac{\pi }{2}+{\alpha }_0=\frac{\pi }{2}+{\sin }^{-1}\frac{R}{a+R}$$$$R^2={SA}^2+{(a+R)}^2-2SA(a+R)\cos \alpha$$$$\Rightarrow {SA}^2-2(a+R)\cos \alpha SA+a(a+2R)=0$$$$SA=(a+R)\cos \alpha \pm \sqrt{{(a+R)}^2{\cos }^2\alpha -a(a+2R)}$$$$SA=(a+R)\{\cos \alpha -\sqrt{{\cos }^2\alpha -\frac{a(a+2R)}{{(a+R)}^2}}\}$$$$SB=(a+R)\{\cos \alpha +\sqrt{{\cos }^2\alpha -\frac{a(a+2R)}{{(a+R)}^2}}\}$$$$AB=SB-SA=2(a+R)\sqrt{{\cos }^2\alpha -\frac{a(a+2R)}{{(a+R)}^2}}$$$$\frac{h-z}{H-h}=\frac{AB}{SA}$$$$\Rightarrow h-z=(H-h)\frac{AB}{SA}=(H-h)\frac{2(a+R)\sqrt{{\cos }^2\alpha -\frac{a(a+2R)}{{(a+R)}^2}}}{(a+R)\{\cos \alpha -\sqrt{{\cos }^2\alpha -\frac{a(a+2R)}{{(a+R)}^2}}\}}$$$$\Rightarrow h-z=\frac{2(H-h)}{\{\frac{1}{\sqrt{1-\frac{a(a+2R)}{{(a+R)}^2}\times \frac{1}{{\cos }^2\alpha }}}-1\}}$$$$\Rightarrow \mathrm{\Delta }z=h-z=\frac{2(H-h)}{\frac{1}{\sqrt{1-\frac{a(a+2R)}{{(a+R)}^2}[1+{\left(\frac{\sin \theta }{\frac{a}{R}+1+\cos \theta }\right)}^2]}}-1}$$$$Illuminatedarea=A$$$$let\lambda =\frac{a}{R}$$$$A=2{\int }_0^{{\theta }_0}\mathrm{\Delta }zRd\theta$$$$\Rightarrow A=4R(H-h){\int }_0^{{\theta }_0}\frac{d\theta }{\frac{1}{\sqrt{1-\frac{a(a+2R)}{{(a+R)}^2}[1+{\left(\frac{\sin \theta }{\frac{a}{R}+1+\cos \theta }\right)}^2]}}-1}$$$$\Rightarrow A=4R(H-h){\int }_0^{\frac{\pi }{2}+{\sin }^{-1}\frac{1}{1+\lambda }}\frac{d\theta }{\frac{1}{\sqrt{1-[1-\frac{1}{{(1+\lambda )}^2}][1+{\left(\frac{\sin \theta }{1+\lambda +\cos \theta }\right)}^2]}}-1}$$$$with\mu =1+\lambda =\frac{a+R}{R}$$$$\frac{A}{4R(H-h)}={\int }_0^{\frac{\pi }{2}+{\sin }^{-1}\frac{1}{\mu }}\frac{d\theta }{\frac{1}{\sqrt{1-[1-\frac{1}{{\mu }^2}][1+{\left(\frac{\sin \theta }{\mu +\cos \theta }\right)}^2]}}-1}$$$$=\frac{1}{{\mu }^2-1}{\int }_0^{\frac{\pi }{2}+{\sin }^{-1}\frac{1}{\mu }}(1+\mu \cos \theta )d\theta$$$$=\frac{1}{{\mu }^2-1}{[\mu \sin \theta +\theta ]}_0^{\frac{\pi }{2}+{\sin }^{-1}\frac{1}{\mu }}$$$$=\frac{1}{{\mu }^2-1}[\mu \sin (\frac{\pi }{2}+{\sin }^{-1}\frac{1}{\mu })+\frac{\pi }{2}+{\sin }^{-1}\frac{1}{\mu }]$$$$=\frac{1}{{\mu }^2-1}[\sqrt{{\mu }^2-1}+\frac{\pi }{2}+{\sin }^{-1}\frac{1}{\mu }]$$$$$$$$\Rightarrow A=\frac{4R(H-h)}{{\mu }^2-1}(\sqrt{{\mu }^2-1}+\frac{\pi }{2}+{\sin }^{-1}\frac{1}{\mu })$$$$with\mu =\frac{a+R}{R}$$

Commented by ajfour last updated on 30/May/19

$$Great,workSir!thankssomuch.$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$

Commented by mr W last updated on 30/May/19

$$thanksforcheckingsir!$$$$theintegralseemstobecomplicated,$$$$but{it}^{\prime }sreallyeasyaftertheexpression$$$$issimplified,seebelow.$$$$\frac{1}{\frac{1}{\sqrt{1-[1-\frac{1}{{\mu }^2}][1+{\left(\frac{\sin \theta }{\mu +\cos \theta }\right)}^2]}}-1}$$$$=\frac{1}{\frac{\mu }{\sqrt{{\mu }^2-({\mu }^2-1)[1+{\left(\frac{\sin \theta }{\mu +\cos \theta }\right)}^2]}}-1}$$$$=\frac{1}{\frac{\mu }{\sqrt{1-({\mu }^2-1){\left(\frac{\sin \theta }{\mu +\cos \theta }\right)}^2}}-1}$$$$=\frac{1}{\frac{(\mu +\cos \theta )\mu }{\sqrt{{(\mu +\cos \theta )}^2-({\mu }^2-1){\sin }^2\theta }}-1}$$$$=\frac{1}{\frac{(\mu +\cos \theta )\mu }{\sqrt{{\mu }^2+2\mu \cos \theta +{\cos }^2\theta -{\mu }^2{\sin }^2\theta +{\sin }^2\theta }}-1}$$$$=\frac{1}{\frac{(\mu +\cos \theta )\mu }{\sqrt{{\mu }^2{\cos }^2\theta +2\mu \cos \theta +1}}-1}$$$$=\frac{1}{\frac{(\mu +\cos \theta )\mu }{\mu \cos \theta +1}-1}$$$$=\frac{\mu \cos \theta +1}{(\mu +\cos \theta )\mu -(\mu \cos \theta +1)}$$$$=\frac{\mu \cos \theta +1}{{\mu }^2-1}$$

Commented by ajfour last updated on 30/May/19

$$TrueSir,thanksagain.$$

Commented by mr W last updated on 30/May/19

$${it}^{\prime }sagainanicequestionsir!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com