Answer to Question #119356 in Calculus for Olivia

Question #119356
Compute ∬D(x+y)dA where D={(x,y)∈R^2:0≤x≤2 and 0≤y≤√(4-x^2)}

.
Select one:
a. 16
b. 8
c. 8/3
d. 16/3
e. 3
1
Expert's answer
2020-06-01T18:35:01-0400

Using Fubini's theorem, we get:

D(x+y)dA=02dx04x2(x+y)dy=02((xy+y22)04x2)dx=\iint_D(x+y)dA=\int_0^2dx\int_{0}^{\sqrt{4-x^2}}(x+y)dy=\int_0^2((xy+\frac{y^2}{2})\biggr\rvert_{0}^{\sqrt{4-x^2}})dx=

=02(x4x2+12(4x2)))dx=12024x2d(4x2)+02(212x2)dx==\int_0^2(x\sqrt{4-x^2}+\frac12(4-x^2)))dx=-\frac12\int_0^2\sqrt{4-x^2}d(4-x^2)+\int_0^2(2-\frac12x^2)dx=

=83+443=163=\frac83+4-\frac43=\frac{16}3

Answer:d). 163\frac{16}3


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