Angelica Fadel
Jan 04, 2025
Evaluate the integral, where R is the triangle with vertices at….?
Evaluate the integral over the region ( R ), which is the triangle with vertices at (-8, 0), (0, 8), and (8, 0). The integral is given by:
[ R \int (2x + y)^2 \, dA ]
Please provide the exact answer in the form of a fraction. I understand that it may be necessary to evaluate two different integrals for each side of the triangle, but I am confused about the process. Could you please help me clarify this? Thank you!
1 Answers
R is bounded by the lines y = 0, x+y = 8, and x-y = -8.
==> x = y-8 to x = 8-y with y in [0, 8].
(So, we only need to write one double integral in this manner.)
So, we obtain
∫(y = 0 to 8) ∫(x = y-8 to 8-y) (2x + y)^2 dx dy
= ∫(y = 0 to 8) (1/6)(2x + y)^3 {for x = y-8 to 8-y} dy
= ∫(y = 0 to 8) (1/6) [(2(8 - y) + y)^3 - (2(y - 8) + y)^3] dy
= ∫(y = 0 to 8) (1/6) [(16 - y)^3 - (3y - 16)^3] dy
= (-1/6) ∫(y = 0 to 8) [(y - 16)^3 + (3y - 16)^3] dy
= (-1/6) [(1/4)(y - 16)^4 + (1/3)(1/4)(3y - 16)^4] {for y = 0 to 8}
= (-1/72) [3(y - 16)^4 + (3y - 16)^4] {for y = 0 to 8}
= 10240/3.
Double check:
http://www.wolframalpha.com/input/?i=%E2%88%AB%28y...
I hope this helps!
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