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Question Number 124976 by liberty last updated on 07/Dec/20
(1) The gravitational force (in lb) of  attraction between two objects is given  by F =(k/x^2 ), where x is the distance  between the objects. If the objects are  10 ft apart, find the work required to  separate them until they are 50 ft apart. Express  the result in terms of k.  (a) (k/(500))      (b) ((2k)/(25))     (c) (k/5)   (d) (k/(40))  (2)One end of a pool is vertical wall 15 ft  wide. What is the force exerted on this  wall by the water if it is 6 ft deep?  The density of water is 62.4 lb/ft^3   (a) 8420 lb    (b) 33,700 lb    (c) 2810 lb  (d) 16,800 lb  (3)Find the area of the surface generated  by revolving the curve about that   indicated axis.  x = 3(√(4−y)) , 0≤y≤((15)/4) , y−axis  (a) (((125)/2)+5(√(10)))π       (b) (((125)/2)−5(√(10)))π  (c) ((125)/2)π                         (d) 5π(√(10))
$$\left(\mathrm{1}\right)\:{The}\:{gravitational}\:{force}\:\left({in}\:{lb}\right)\:{of} \\ $$$${attraction}\:{between}\:{two}\:{objects}\:{is}\:{given} \\ $$$${by}\:{F}\:=\frac{{k}}{{x}^{\mathrm{2}} },\:{where}\:{x}\:{is}\:{the}\:{distance} \\ $$$${between}\:{the}\:{objects}.\:{If}\:{the}\:{objects}\:{are} \\ $$$$\mathrm{10}\:{ft}\:{apart},\:{find}\:{the}\:{work}\:{required}\:{to} \\ $$$${separate}\:{them}\:{until}\:{they}\:{are}\:\mathrm{50}\:{ft}\:{apart}.\:{Express} \\ $$$${the}\:{result}\:{in}\:{terms}\:{of}\:{k}. \\ $$$$\left({a}\right)\:\frac{{k}}{\mathrm{500}}\:\:\:\:\:\:\left({b}\right)\:\frac{\mathrm{2}{k}}{\mathrm{25}}\:\:\:\:\:\left({c}\right)\:\frac{{k}}{\mathrm{5}}\:\:\:\left({d}\right)\:\frac{{k}}{\mathrm{40}} \\ $$$$\left(\mathrm{2}\right){One}\:{end}\:{of}\:{a}\:{pool}\:{is}\:{vertical}\:{wall}\:\mathrm{15}\:{ft} \\ $$$${wide}.\:{What}\:{is}\:{the}\:{force}\:{exerted}\:{on}\:{this} \\ $$$${wall}\:{by}\:{the}\:{water}\:{if}\:{it}\:{is}\:\mathrm{6}\:{ft}\:{deep}? \\ $$$${The}\:{density}\:{of}\:{water}\:{is}\:\mathrm{62}.\mathrm{4}\:{lb}/{ft}^{\mathrm{3}} \\ $$$$\left({a}\right)\:\mathrm{8420}\:{lb}\:\:\:\:\left({b}\right)\:\mathrm{33},\mathrm{700}\:{lb}\:\:\:\:\left({c}\right)\:\mathrm{2810}\:{lb}\:\:\left({d}\right)\:\mathrm{16},\mathrm{800}\:{lb} \\ $$$$\left(\mathrm{3}\right){Find}\:{the}\:{area}\:{of}\:{the}\:{surface}\:{generated} \\ $$$${by}\:{revolving}\:{the}\:{curve}\:{about}\:{that}\: \\ $$$${indicated}\:{axis}.\:\:{x}\:=\:\mathrm{3}\sqrt{\mathrm{4}−{y}}\:,\:\mathrm{0}\leqslant{y}\leqslant\frac{\mathrm{15}}{\mathrm{4}}\:,\:{y}−{axis} \\ $$$$\left({a}\right)\:\left(\frac{\mathrm{125}}{\mathrm{2}}+\mathrm{5}\sqrt{\mathrm{10}}\right)\pi\:\:\:\:\:\:\:\left({b}\right)\:\left(\frac{\mathrm{125}}{\mathrm{2}}−\mathrm{5}\sqrt{\mathrm{10}}\right)\pi \\ $$$$\left({c}\right)\:\frac{\mathrm{125}}{\mathrm{2}}\pi\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({d}\right)\:\mathrm{5}\pi\sqrt{\mathrm{10}}\: \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 07/Dec/20
Work done =∫_(10) ^(50) (k/x^2 )dx=[−(k/x)]_(10) ^(50) =((2k)/(25))
$${Work}\:{done}\:=\int_{\mathrm{10}} ^{\mathrm{50}} \frac{{k}}{{x}^{\mathrm{2}} }{dx}=\left[−\frac{{k}}{{x}}\right]_{\mathrm{10}} ^{\mathrm{50}} =\frac{\mathrm{2}{k}}{\mathrm{25}} \\ $$
Answered by bemath last updated on 07/Dec/20
(3)S = ∫_0 ^(15/4) 2πx (√(1+((dx/dy))^2 )) dy  x = 3(√(4−y)) ; (dx/dy) = ((−3)/(2(√(4−y))))   then 1+((dx/dy))^2 =1+(9/(4(4−y)))=((25−4y)/(4(4−y)))  S = ∫_0 ^( ((15)/4))  6π(√(4−y)) (√((25−4y)/(4(4−y)))) dy  S= 3π∫_0 ^((15)/4) (√(25−4y))  dy = [(1/((3/2)(−4)))×3π×(√((25−4y)^3 )) ]_0 ^((15)/4)   = −(π/2) {(√(1000))−125 }  = ((125π)/2)−5π(√(10)) = (((125)/2)−5(√(10)))π
$$\left(\mathrm{3}\right){S}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{15}/\mathrm{4}} {\int}}\mathrm{2}\pi{x}\:\sqrt{\mathrm{1}+\left(\frac{{dx}}{{dy}}\right)^{\mathrm{2}} }\:{dy} \\ $$$${x}\:=\:\mathrm{3}\sqrt{\mathrm{4}−{y}}\:;\:\frac{{dx}}{{dy}}\:=\:\frac{−\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{4}−{y}}} \\ $$$$\:{then}\:\mathrm{1}+\left(\frac{{dx}}{{dy}}\right)^{\mathrm{2}} =\mathrm{1}+\frac{\mathrm{9}}{\mathrm{4}\left(\mathrm{4}−{y}\right)}=\frac{\mathrm{25}−\mathrm{4}{y}}{\mathrm{4}\left(\mathrm{4}−{y}\right)} \\ $$$${S}\:=\:\underset{\mathrm{0}} {\overset{\:\frac{\mathrm{15}}{\mathrm{4}}} {\int}}\:\mathrm{6}\pi\sqrt{\mathrm{4}−{y}}\:\sqrt{\frac{\mathrm{25}−\mathrm{4}{y}}{\mathrm{4}\left(\mathrm{4}−{y}\right)}}\:{dy} \\ $$$${S}=\:\mathrm{3}\pi\underset{\mathrm{0}} {\overset{\frac{\mathrm{15}}{\mathrm{4}}} {\int}}\sqrt{\mathrm{25}−\mathrm{4}{y}}\:\:{dy}\:=\:\left[\frac{\mathrm{1}}{\frac{\mathrm{3}}{\mathrm{2}}\left(−\mathrm{4}\right)}×\mathrm{3}\pi×\sqrt{\left(\mathrm{25}−\mathrm{4}{y}\right)^{\mathrm{3}} }\:\right]_{\mathrm{0}} ^{\frac{\mathrm{15}}{\mathrm{4}}} \\ $$$$=\:−\frac{\pi}{\mathrm{2}}\:\left\{\sqrt{\mathrm{1000}}−\mathrm{125}\:\right\} \\ $$$$=\:\frac{\mathrm{125}\pi}{\mathrm{2}}−\mathrm{5}\pi\sqrt{\mathrm{10}}\:=\:\left(\frac{\mathrm{125}}{\mathrm{2}}−\mathrm{5}\sqrt{\mathrm{10}}\right)\pi \\ $$
Answered by bemath last updated on 07/Dec/20
(2)Force = whA = 62.4 lb/ft^3  ×6 ft ×15^2  ft^2    = 84,240 lb
$$\left(\mathrm{2}\right){Force}\:=\:{whA}\:=\:\mathrm{62}.\mathrm{4}\:{lb}/{ft}^{\mathrm{3}} \:×\mathrm{6}\:{ft}\:×\mathrm{15}^{\mathrm{2}} \:{ft}^{\mathrm{2}} \\ $$$$\:=\:\mathrm{84},\mathrm{240}\:{lb} \\ $$

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