How do you find #\int _ { - 1} ^ { 1} \int _ { 3} ^ { 4}\int _ { 0} ^ { 2} ( x y ^ { 2} + y z ^ { 2} ) d z d y d x#?

1 Answer
Steve M
Dec 24, 2016

# int_{-1}^{1} int_{3}^{4} int_{0}^{2} ( xy^2 + yz^2) \ dz \ dy \ dx = 56/3#

Explanation:

We want to evaluate:

# int_{-1}^{1} int_{3}^{4} int_{0}^{2} ( xy^2 + yz^2) \ dz \ dy \ dx#

We perform multiple integration by starting with the inner integral and integrate wrt to the variable of integration whilst treating other variables as constant. So:

# int_{-1}^{1} int_{3}^{4} int_{0}^{2} (xy^2 + yz^2) \ dz \ dy \ dx #
# " "= int_{-1}^{1} int_{3}^{4} [xy^2z + (yz^3)/3]_(z=0)^(z=2) \ dy \ dx #

# " "= int_{-1}^{1} int_{3}^{4} {(2xy^2 + (8y)/3)-(0+0)} \ dy \ dx #
# " "= int_{-1}^{1} int_{3}^{4} (2xy^2 + (8y)/3) \ dy \ dx #

# " "= int_{-1}^{1} [(2xy^3)/3 + (8y^2)/6]_{y=3}^{y=4} \ dx #

# " "= int_{-1}^{1} 1/3[2xy^3 + 4y^2]_{y=3}^{y=4} \ dx #

# " "= int_{-1}^{1} 1/3{(2x*64) + (4*16))-((2x*27) + (4*9))} \ dx #

# " "= int_{-1}^{1} 1/3{128x + 64)-(54x + 36)} \ dx #

# " "= int_{-1}^{1} 1/3(74x + 28) \ dx #
# " "= 1/3int_{-1}^{1} (74x + 28) \ dx #

# " "= 1/3 [(74x^2)/2 + 28x]_{x=-1}^{x=1} \ dx #

# " "= 1/3 [37x^2 + 28x]_{x=-1}^{x=1} \ dx #

# " "= 1/3 {(37 + 28)-(37 - 28)} #
# " "= 1/3 {65-9} #
# " "= 56/3 #

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