document.write( "Question 1158831: the area bounded by the curves y=(x^3)/4, x+y=4 and y=0 is to be revolved about the y=axis. Find the volume generated. \n" ); document.write( "

Algebra.Com's Answer #781824 by math_helper(2461) $\"\"$ $\"About$

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document.write( "The region described can be divided into two sections.  The first section is from x=0 to x=2  and is the area under y=(1/4)x^3.  Note that  y=-x+4 and y=(1/4)x^3  intersect at (2,2).  The second section is from x=2 to x=4 and is simply the area under y=-x+4.\r
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document.write( "Using concentric cylinders...\r
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document.write( "Each cylinder has an infinitesimal volume that is equal to 2pi*r*h*dx.  If it helps, you can envision one such cylinder as cut and rolled out to form a rectangle.  That rectangle has height equal to that of the function 'h' ((1/4)x^3, for the first section) and 'r' is just x so the length is 2*pi*x.   Finally, for the 3rd dimension, the thickness is dx.  \r
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document.write( "For the first section then:\r
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document.write( "And by integration, we can add all these infinitesimal volumes, allowing for the varying heights of the cylinders as f(x) changes over the limits of integration:\r
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document.write( " evaluated at x=2 minus value at x=0\r
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document.write( "And the 2nd section... ( x=2 to x=4 ):\r
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document.write( " evaluated at x=4 minus value at x=2
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document.write( "   (approx. 43.56 cubic units)\r
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document.write( "I think this is right. Its late though, and for me its always dangerous doing math when tired...
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