How do you solve #\int _ { 0} ^ { 2\pi } \int _ { 0} ^ { 6} 3r ^ { 2} \sin \theta d r d \theta#?

1 Answer
Steve M
Nov 27, 2016

# I = int_0^(2pi) int_0^6 3r^2 sin theta dr d theta = 0#

Explanation:

Let # I = int_0^(2pi) int_0^6 3r^2 sin theta dr d theta #, Then:

# I = int_0^(2pi) (int_0^6 3r^2 sin theta dr )d theta #
# :. I = int_0^(2pi) [3r^3/3sin theta]_(r=0)^(r=6) d theta #
# :. I = int_0^(2pi) sin theta[r^3 ]_(r=0)^(r=6) d theta #
# :. I = int_0^(2pi) sin theta(216-0) d theta #
# :. I = int_0^(2pi) (216sin theta) d theta #
# :. I = [-216cos theta]_0^(2pi) #
# :. I = -216[cos theta]_0^(2pi) #
# :. I = -216(1-1) #
# :. I = 0 #

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