document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1205078>Question 1205078</A>: <I><SPAN STYLE=\"word-break: break-all;\"> From a cylindrical object of diameter 70cm and height 84cm, a right solid cone having its base as one of the circular ends of the cylinder and height 84cm is removed.<BR>\n" );
document.write( "Calculate:<BR>\n" );
document.write( "a) The volume of the remaining solid object, expressing your answer in the form of a × 10ⁿ where 1 < a < 10 and n is a positive integer.<BR>\n" );
document.write( "b) The surface area of the remaining solid object </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #841710 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=841710\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Part (a)\r<BR>\n" );
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document.write( "If a cylinder and cone have the same radius and height, then we have this very interesting connection:<BR>\n" );
document.write( "cone = (1/3)*cylinder<BR>\n" );
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document.write( "cylinder = 3*cone\r<BR>\n" );
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document.write( "This is an informal way of saying \"we need 3 cones to make a cylinder\". <BR>\n" );
document.write( "In other words, 3 cone volumes combine to a cylinder volume.<BR>\n" );
document.write( "Again, both must share the same radius and same height.\r<BR>\n" );
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document.write( "Luckily the radius values are the same because we're carving a cone out of the cylinder. <BR>\n" );
document.write( "And both heights are the same as well (84).\r<BR>\n" );
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document.write( "After carving a cone out of the cylinder, the cylinder loses 1/3 of its volume and keeps the remaining 2/3.\r<BR>\n" );
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document.write( "volume of cylinder = pi*r^2*h<BR>\n" );
document.write( "volume of cylinder = pi*(70/2)^2*84<BR>\n" );
document.write( "volume of cylinder = 102900pi\r<BR>\n" );
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document.write( "2/3 of that volume is (2/3)*102900pi = 68600pi which is the leftover amount.\r<BR>\n" );
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document.write( "Your teacher hasn't stated something like \"use pi = 3.14\", so I'll use the calculator's stored version of pi instead.<BR>\n" );
document.write( "68600pi = 215,513.25603626<BR>\n" );
document.write( "which rounds to 215,513\r<BR>\n" );
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document.write( "That converts to the scientific notation <font color=red>2.15513 * 10^5</font><BR>\n" );
document.write( "This is because we move the decimal point 5 spots to the right to go from 2.15513 back to 215,513 again.\r<BR>\n" );
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document.write( "Part (b)\r<BR>\n" );
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document.write( "The cylinder has a circular floor area of: <BR>\n" );
document.write( "pi*r^2<BR>\n" );
document.write( "= pi*(70/2)^2<BR>\n" );
document.write( "= 1225pi<BR>\n" );
document.write( "We'll use this value later. Let x = 1225pi\r<BR>\n" );
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document.write( "The area of the curved walls, aka lateral area of the cylinder, is:<BR>\n" );
document.write( "(circumference of circular base)*(height)<BR>\n" );
document.write( "= (pi*diameter)*(height)<BR>\n" );
document.write( "= (pi*70)*(84)<BR>\n" );
document.write( "= 5880pi<BR>\n" );
document.write( "We'll use this value later. Let y = 5880pi\r<BR>\n" );
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document.write( "Then the last piece to consider is the lateral surface area of the cone carved out of the cylinder. <BR>\n" );
document.write( "We can think of this as an \"inverted\" surface area of sorts, since we're effectively looking at the inside wall of the cone we carved out. \r<BR>\n" );
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document.write( "The lateral surface area of a cone is<BR>\n" );
document.write( "pi*r*L<BR>\n" );
document.write( "where<BR>\n" );
document.write( "L = slant height of the cone<BR>\n" );
document.write( "L = sqrt(r^2+h^2) due to the pythagorean theorem\r<BR>\n" );
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document.write( "We can update that formula to get<BR>\n" );
document.write( "pi*r*sqrt(r^2+h^2)<BR>\n" );
document.write( "= pi*(70/2)*sqrt((70/2)^2+84^2)<BR>\n" );
document.write( "= 3185pi<BR>\n" );
document.write( "We'll use this value later. Let z = 3185pi\r<BR>\n" );
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document.write( "The last thing to do is add the results we got to determine the entire surface area of this strange 3D shape.<BR>\n" );
document.write( "x+y+z = 1225pi+5880pi+3185pi <BR>\n" );
document.write( "= (1225+5880+3185)pi<BR>\n" );
document.write( "= 10290pi\r<BR>\n" );
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document.write( "The total exact surface area of this strange 3D shape is <font color=red>10290pi square cm</font>\r<BR>\n" );
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document.write( "I'll let the student compute the approximate version of this value. <BR>\n" );
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