Asked by Brandon
I'm having trouble reversing the order of integration of ∫∫dxdy from a=0 to b=2(3)^(1/2) for x and c=y^(2/6) to d=(16-y^2)^(1/2) for y.
I graphed the region of integration and that still doesn't really help me.
i got approximately 7.9 for the ∫∫dxdy which im pretty sure is right.
please help
I graphed the region of integration and that still doesn't really help me.
i got approximately 7.9 for the ∫∫dxdy which im pretty sure is right.
please help
Answers
Answered by
Steve
I'm having trouble interpreting your variables and limits.
c,d are the limits on dy? That's odd, since I'd expect functions of y to be limits on dx. Anyway, interpreting it as
∫[0,√12]∫[y^(1/3),sqrt(16-y^2)] dx dy I get 15.8, twice your answer
I tried to get a handle on the region using the excellent graphing tools at rechneronline dot de slash function-graphs, and it appears the region is to the right of the curve y=x^3 and inside the circle y=sqrt(16-x^2) for x in [0,√12]. Strange limits, since they don't appear to correspond to any useful portion of the picture.
??
c,d are the limits on dy? That's odd, since I'd expect functions of y to be limits on dx. Anyway, interpreting it as
∫[0,√12]∫[y^(1/3),sqrt(16-y^2)] dx dy I get 15.8, twice your answer
I tried to get a handle on the region using the excellent graphing tools at rechneronline dot de slash function-graphs, and it appears the region is to the right of the curve y=x^3 and inside the circle y=sqrt(16-x^2) for x in [0,√12]. Strange limits, since they don't appear to correspond to any useful portion of the picture.
??
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