• anonymous
I've tried solving this problem for over an hour. Just cant seem to get it. Any help is much appreciated!! Find the volume of the solid that results when the region enclosed by the curves y=(x^2)/81, x=9y^2 and is revolved about the y-axis
  • Stacey Warren - Expert
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  • chestercat
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  • anonymous
first of all you have to find intersection points between two curves .. if we draw it .. it will be so similar to this : |dw:1328516145242:dw| \[81y-x^{2}= 0...................1\] \[-9y^{2}+x=0.......................2\] \[y=\frac{x^{2}}{81}....................1\] sub last eq in 2 \[-9\left( \frac{x^{2}}{81} \right)^{2}+x=0\] \[- \cancel {9}\left( \frac{x^{4}}{\cancel{81}\times81} \right)+x=0\] \[-x^{4}+(9\times81)x=0\] Take x as combination and it will give us option of x=0 \[-x^{3}+(81\times9) = 0\] x= 9 then this indicate to us the interval of integration that is being from zero to 9 and the equation used that in x parameter only which is : \[-x^{4}+(9\times81)x=0\] \[\int\limits_{9}^{0}(-x^{4}+(9\times81)x)dx=(-\frac{x5}{5}+\frac{(9\times81)x^2}{2})_{0}^{9}\] \[(-\frac{9^{5}}{5}+\frac{(9 \times 81)9^{2})}{2})-\cancel{(-\frac{0^{5}}{5}+\frac{(9 \times 81))0^{2})}{2})}\] =-8529.3 unit area by the way .. should write this no. with out -ve sign and i insist to show it to you .. the equation that we dependent it in integral ... we have to multilpliy it by -ve sign as long as it equal to zero ..and nothing changed.. its also make the result +ve.. thnx for your Q. i really enjoy solute it > regards

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