Use double integration to calculate the area that liew between the curves This is what i have done , between the limits on the y = a-x and y = 0 and for x = a and 0 But according to my text book the answer is
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Originally Posted by adam_leeds Use double integration to calculate the area that liew between the curves This is what i have done , between the limits on the y = a-x and y = 0 and for x = a and 0 But according to my text book the answer is Your limits of integration are incorrect. If you sketch the region you will see that the line lies above the curves and they both have x intercept This gives
Where did you get y = 0? Why not use the first equation? We probably should establish, due ot the first constraint, that a >= 0.
Originally Posted by TheEmptySet This gives That doesnt get a^2/3
Originally Posted by adam_leeds That doesnt get a^2/3 ummm no. This gives
Missed a negative, thanks.
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