- #1
coderdave
- 5
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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
\int_{ }^{ } \int_{ }^{ } z^2 dS = \int_{ }^{ } \int_{ }^{ } \sqrt[ ]{ 1 - x^2 - y^2 } dx dy
"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
\int_{0}^{ \pi/2} \int_{0}^{1} \sqrt[ ]{ 1 - r^2cos^2( \theta) - r^2sin^2(\theta) } r dr d\theta
Any ideas getting past this point?
\int_{ }^{ } \int_{ }^{ } z^2 dS = \int_{ }^{ } \int_{ }^{ } \sqrt[ ]{ 1 - x^2 - y^2 } dx dy
"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
\int_{0}^{ \pi/2} \int_{0}^{1} \sqrt[ ]{ 1 - r^2cos^2( \theta) - r^2sin^2(\theta) } r dr d\theta
Any ideas getting past this point?
