- #1
kira137
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Homework Statement
Evaluate using the cylindrical shell method, the volume of the solid of revolution obtained by rotating the region enclosed by the curves y=x^3, y=-x and y=1 around the line y=-1.
The attempt at a solution
The answer is suppose to be 127(pi)/42 but I keep getting 169(Pi)/42.
I first changed y=x^3 to x= y^(1/3) and y=-x to x=-y
then found radius = y+1
then height as y^(1/3)+y
then plugged into cylindrical formula.
1
S 2(pi)(y+1)(y^(1/3)+y) dy =
0
,,,,,,,1
2(pi) S y^(4/3)+y^2+y^(1/3)+y dy =
,,,,,,,0
2(pi) [(y^(7/3)/7/3) + (y^3)/3 + ((y^4/3)/4/3) +(y^2)/2] 0 to 1 =
2(pi) [3/7 + 1/3 + 3/4 + 1/2] =
2(pi) [36/84 + 28/84 + 63/84 + 42/84] =
2(pi)(169/84) = 338(pi)/84 = 169(pi)/42
I'm not sure where I went wrong.. I went over and over but I couldn't find out the problem.
your help is greatly appreciated, thank you!
Evaluate using the cylindrical shell method, the volume of the solid of revolution obtained by rotating the region enclosed by the curves y=x^3, y=-x and y=1 around the line y=-1.
The attempt at a solution
The answer is suppose to be 127(pi)/42 but I keep getting 169(Pi)/42.
I first changed y=x^3 to x= y^(1/3) and y=-x to x=-y
then found radius = y+1
then height as y^(1/3)+y
then plugged into cylindrical formula.
1
S 2(pi)(y+1)(y^(1/3)+y) dy =
0
,,,,,,,1
2(pi) S y^(4/3)+y^2+y^(1/3)+y dy =
,,,,,,,0
2(pi) [(y^(7/3)/7/3) + (y^3)/3 + ((y^4/3)/4/3) +(y^2)/2] 0 to 1 =
2(pi) [3/7 + 1/3 + 3/4 + 1/2] =
2(pi) [36/84 + 28/84 + 63/84 + 42/84] =
2(pi)(169/84) = 338(pi)/84 = 169(pi)/42
I'm not sure where I went wrong.. I went over and over but I couldn't find out the problem.
your help is greatly appreciated, thank you!