How do you find the volume of the solid bounded by Z = 1 – y^2, x + y = 1, and the three coordinate plane?

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Answer 1 · 2016-08-11T08:39:39

#5/12#

plugging the limits into #V = int_V \ dV = int_V dx \ dy \ dz#, we have

#V = int_(x = 0)^1 int_(y=0)^(1-x) int_(z = 0)^(1 - y^2) dz \ dy \ dx#

# = int_(x = 0)^1 int_(y=0)^(1-x) 1 - y^2 \ dy \ dx#

# = int_(x = 0)^1 [ y - y^3/3 ]_(0)^(1-x) \ dx#

# = int_(x = 0)^1 1-x -(1-x)^3/3 \ dx#

# = [ x-x^2/2 + (1-x)^4/12 ]_0^1 #

# = [ 1-1 /2 ] - [1/12] = 5/12#