Find the surface area of the solid formed when x=(4y^1/3)-((1/12)y^1/2) for yE[1,4] rotated around the y-axis.  Step by step would be great.  I am having trouble understanding how to do this.
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Consider x and y interchanged: y=(4x^(1/3))-((1/12)x^(1/2)) for limits of x between 1 and 4 rotated about the x axis. The volume will be just the same. When x=0, y=0 so the curve passes through the origin. When x=1, y=47/12 or about 4. When x=64 (for example), y=16-8/12=16-2/3=46/3 or about 15. When x=729, y=36-27/12=36-9/4=135/4 or about 34. Sketching these points gives you a better picture of what you are going for. The curve rises from zero to about 4 as x goes from 0 to 1 then rises only very gradually as x increases further. You need to consider the bit between x=1 and 4. Consider this part of the curve as a set of very thin discs with the x axis as a spindle through the centre of the discs. The radius of each disc is the y value and dx is the thickness. The volume of each disc is (pi)y^2dx. This is the basis of the integral. Replace y^2 by squaring the function. 

y^2=16x^(2/3)-(8/12)x^(5/6)+(1/144)x

and integrate wrt x: (3/5)16x^(5/3)-(6/11)(8/12)x^(11/6)+(1/288)x^2.

Put x=4 (upper limit): (3/5)16*4^(5/3)-(4/11)4^(11/6)+(1/288)16=92.2 approx.

Put x=1 (lower limit): (3/5)16-(4/11)+(1/288)=9.24 approx.

Subtract: 92.2-9.24=82.96 approx. and multiply by (pi)=260.63 approx. That's the volume.

(You could have used (pi)x^2dy as the integral without interchanging x and y)

by Top Rated User (1.0m points)

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