I'm self learning and working my way through the book "Div Grad Curl and All That". On one of the pages (27) the author says(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \int_{ }^{ } \int_{ }^{ } z^2 dS = \int_{ }^{ } \int_{ }^{ } \sqrt[ ]{ 1 - x^2 - y^2 } dx dy[/tex]

"This is an ordinary double integral, and you should verify that its value is pi / 6.

The hint was to convert to polar coordinates and the furthest I got was

[tex] \int_{0}^{ \pi/2} \int_{0}^{1} \sqrt[ ]{ 1 - r^2cos^2( \theta) - r^2sin^2(\theta) } r dr d\theta[/tex]

Any ideas getting past this point?

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

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

# Having trouble with a surface integral

Loading...

Similar Threads - Having trouble surface | Date |
---|---|

Having trouble understanding this integration | Dec 19, 2013 |

Having some trouble understanding a u-substitution | Jul 14, 2013 |

I am having trouble with 2 volume and a profit problems? | Nov 21, 2012 |

I'm having trouble with an integral for my QM senior thesis | Aug 1, 2012 |

Trouble proving a function with 2 variables does not have a global max or global min | Jul 26, 2012 |

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