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rim182946
 3 years ago
Best ResponseYou've already chosen the best response.1I think you need more information like z = 0 ( so it can be above the xy plane )

AbdulRehman
 3 years ago
Best ResponseYou've already chosen the best response.0Then find the area only in first ocatne

rim182946
 3 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{0}^{\pi/2}\int\limits_{0}^{2}\int\limits_{0}^{r \cos \theta+4r \sin \theta}rdzdrd \theta\]

rim182946
 3 years ago
Best ResponseYou've already chosen the best response.1you should use cyindrical because the region in the xy plane is circular in nature you have to convert the z bounds then the low z bound is z = 0 and the high z bound is z = x + 4y you said only the first octant so after the z integration you are down to the quarter circle in the first quadrant which has r bounds r = 0 to r = 2 (the circle of radius 2 centered at the origin) theta bounds of theta = o to theta = Pi/2

AbdulRehman
 3 years ago
Best ResponseYou've already chosen the best response.0Well that is quite right. I find difficulty in drawing graph in 3d through and then finding it's limit. Any guess.
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