the question is..(adsbygoogle = window.adsbygoogle || []).push({});

Find the areas of the surfaces generated by revolvin the curves about the indicated axes.

x = (1/3)y^(3/2) - y^(1/2), 1≦y≦3; revolved about y-axis

so i use the general formula "S = Integral 2π (radius)(dS)"

and the radiu in this case is x which is (1/3)y^(3/2) - y^(1/2) but yea..

i just keep getting negative value when i subsitute everything in the equation...

thanks.

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Can we get a negative value for areas of surfaces of revolution?

Loading...

Similar Threads - negative value areas | Date |
---|---|

I Q about finding area with double/volume with triple integral | Sep 13, 2017 |

B Definite integrals with +ve and -ve values | Jun 10, 2017 |

Simpson's Rule with negative values | Aug 19, 2014 |

Ignoring positive/negative values with trig substitutions? | Sep 1, 2011 |

Negative values in covariance matrix | Aug 10, 2008 |

**Physics Forums - The Fusion of Science and Community**