Question

Please show me where I went wrong with solving the following problem (click to see calculations). Find the volume of the solid that results when the region enclosed by x=y^2 and x=y is revolved about the line y=-1

Answer 1

\[\pi \int\limits_{0}^{1}[(\sqrt{x}-(-1))^2-(x-(-1))^2]dx\]
\[=\pi \int\limits_{0}^{1}[x-x^2]dx=\pi[\frac{ x^2 }{ 2 }-\frac{ x^3 }{ 3 }]^{1}_{0}\]

Answer 2

\[=\frac{ \pi }{ 6 }\]

Answer 3

the answer in the book is pi/2

Answer 4

@zepdrix

Answer 5

you can still edit your question

Answer 6

thanks

Answer 7

bare with me just checking your math from your set up