How do you find?#int_0^2int_0^sqrt(4-x^2)(4-y^2)^(3/2)dydx#

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#int_0^2int_0^sqrt(4-x^2)(4-y^2)^(3/2)dydx#

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Answer 1 · 2018-06-24T13:08:33

#= 256/15 #

#int_(x=0)^2int_(y=0)^sqrt(4-x^2)(4-y^2)^(3/2)dydx#

Switch the order

#= int_(y = 0)^2 int_(x=0)^sqrt(4- y^2)(4-y^2)^(3/2) dx dy#

#= int_(y = 0)^2 [ x (4-y^2)^(3/2)]_(x=0)^sqrt(4- y^2) dy#

#= int_(y = 0)^2 (4-y^2)^(2) dy#

#= [16y- 8/3 y^3 + 1/5 y^5 ]_(y = 0)^2#

#= 32- 64/3 + 32/5 = 256/15 #