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Question Number 60088 by ajfour last updated on 17/May/19

Commented by ajfour last updated on 17/May/19

$$\mathrm{Only}\mathrm{hemisphere}\mathrm{of}\mathrm{radius}\mathrm{R}\mathrm{is}\mathrm{grassy},$$$$\mathrm{and}\mathrm{occupies}\mathrm{central}\mathrm{area}\mathrm{of}\mathrm{a}$$$$\mathrm{square}\mathrm{field}\mathrm{of}\mathrm{side}\mathrm{a}>2\mathrm{R}.$$$$\mathrm{A}\mathrm{goat}\mathrm{tied}\mathrm{at}\mathrm{front}\mathrm{right}\mathrm{corner}$$$$\mathrm{with}\mathrm{a}\mathrm{rope}\mathrm{of}\mathrm{length}l=\frac{a}{2},$$$$\mathrm{can}\mathrm{graze}\max .\mathrm{what}\mathrm{area}\mathrm{of}\mathrm{grass}?$$$$(\mathrm{express}\mathrm{answer}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{a},\mathrm{and}\mathrm{R}.$$

Answered by mr W last updated on 17/May/19

Commented by mr W last updated on 17/May/19

$$d=\frac{\sqrt{2}a}{2}$$$$\sqrt{R^2-r^2}=d-\sqrt{l^2-r^2}$$$$R^2-r^2=d^2+l^2-r^2-2d\sqrt{l^2-r^2}$$$$d^2+l^2-R^2=2d\sqrt{l^2-r^2}$$$${(d^2+l^2-R^2)}^2=4d^2(l^2-r^2)$$$$r=\sqrt{l^2-{\left(\frac{d^2+l^2-R^2}{2d}\right)}^2}$$$$surfaceareaofsphericalcap:$$$$A_{cap}=2\pi R(R-\sqrt{R^2-r^2})$$$$duetohemispherethegrassareais$$$$halfofthesphericalcap:$$$$A_g=\pi R(R-\sqrt{R^2-r^2})=\pi R[R-\sqrt{R^2-l^2+{\left(\frac{d^2+l^2-R^2}{2d}\right)}^2}]$$$$=\pi R[R-\sqrt{R^2-l^2+\frac{1}{8}{\left(\frac{a^2+2l^2-2R^2}{a}\right)}^2}]$$

Commented by ajfour last updated on 17/May/19

$$\mathrm{Hope}\mathrm{i}\mathrm{shall}\mathrm{follow}..$$$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{amazingly}\mathrm{quick}\mathrm{my}\mathrm{Sir}!$$

Commented by ajfour last updated on 17/May/19

$$A_{cap}=2\pi R(R-\sqrt{R^2-r^2})$$$$(\mathrm{Sir}\mathrm{please}\mathrm{explain}{\mathrm{this}}^{\prime }\mathrm{wont}\mathrm{we}$$$$\mathrm{need}\mathrm{to}\mathrm{integrate}?)$$

Commented by mr W last updated on 17/May/19

$$surfaceareaofasphericalcapwith$$$$heighthis2\pi Rh.iusedthisdirectly.$$

Commented by mr W last updated on 17/May/19

Commented by otchereabdullai@gmail.com last updated on 17/May/19

$$\mathrm{I}\mathrm{salute}\mathrm{you}\mathrm{prof}\mathrm{W}$$

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