document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=2177>Question 2177</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Dear sir,<BR>\n" );
document.write( "It is not a hollow cylinder .It is a  verticallly positioned rotating cylinder(i.e. cylinder in sleeping position ,or in other words,  the cylinder height is parallel to the earth suface).It is filled with balls (of different dimensions) to a particular height inside the mill. Company people will not give the amount of balls in the mill.they will provide only the height upto which the balls are present inside the mill.they will not give other data except this. <BR>\n" );
document.write( "So my data are follows,<BR>\n" );
document.write( "Mill dia = 2.21m<BR>\n" );
document.write( "Mill Length = 3.21m<BR>\n" );
document.write( "Height upto which the balls are present inside the mill = 1.14m<BR>\n" );
document.write( "SO the free space inside the mill is (2.21(dia) - 1.14) = 1.07 m<BR>\n" );
document.write( "SO I have considered this part as a segment area. Hope now you can understand the concept.<BR>\n" );
document.write( "Please reply the various formulae for Segment Area soon<BR>\n" );
document.write( "Thankyou,<BR>\n" );
document.write( "Prem\r<BR>\n" );
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document.write( "  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #876 by </B><A HREF=/tutors/aboutme.mpl?userid=longjonsilver><B>longjonsilver(2297)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=longjonsilver><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=longjonsilver><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=876\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> ok, got it :-)\r<BR>\n" );
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document.write( "Let centre of circle be O, Let height of balls be a horizontal line AB, which is 1.14m above the bottom of the cyclinder (and is above the centre of the circle) since cylinder radius = 1.105m.\r<BR>\n" );
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document.write( "I shall consider the free space above the balls...this is the segment made by the line AB, as part of the sector AOB. Given that the angle AOB is a (in radians!!), then we know the following formulae:\r<BR>\n" );
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document.write( "area of sector AOB = (1/2)r^2a<BR>\n" );
document.write( "area of triangle AOB = (1/2)r^2sina\r<BR>\n" );
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document.write( "so area of segment made by AB is area of sector - area of triangle<BR>\n" );
document.write( "segment = (1/2)r^2a - (1/2)r^2*sina<BR>\n" );
document.write( "segment = (1/2)r^2(a-sina)\r<BR>\n" );
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document.write( "We need the angle a!\r<BR>\n" );
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document.write( "Draw a vertical line up from O to the mid-point of AB, call this point C. Now look at the right angled triangle AOC....\r<BR>\n" );
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document.write( "OA = radius = 1.105m<BR>\n" );
document.write( "OC = 1.14-1.105m = 0.035m<BR>\n" );
document.write( "angle AOC = b\r<BR>\n" );
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document.write( "so cosb=OC/OA<BR>\n" );
document.write( "therefore b = 88.18489degrees\r<BR>\n" );
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document.write( "so angle a = 2b = 176.3697 degrees.\r<BR>\n" );
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document.write( "Now to convert this to radians....\r<BR>\n" );
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document.write( "360 degree = 2pi radians<BR>\n" );
document.write( "so 1 degree is (2pi)/360 radians<BR>\n" );
document.write( "hence 176.36 degrees is 3.07823 radians (to however many dp you want)\r<BR>\n" );
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document.write( "so,\r<BR>\n" );
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document.write( "segment above balls = (1/2)(1.105)^2(3.07823-sin(3.07823))<BR>\n" );
document.write( " = 1.8406m^2.\r<BR>\n" );
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document.write( "As cylinder area = pi*r^2<BR>\n" );
document.write( "total Area = 3.83596m^2\r<BR>\n" );
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document.write( "then area of balls = 3.83596-1.8406 = 1.99546m^2\r<BR>\n" );
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document.write( "Hope this helps...Please check my working though, as i was rushing (i am at work), but the approach is correct.\r<BR>\n" );
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document.write( "jon.\r<BR>\n" );
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