Here's the question you clicked on:
AbdulRehman
find the volume enclosed by x²+y²=4 and z=x+4y using multiple integrals?
I think you need more information like z = 0 ( so it can be above the xy plane )
Then find the area only in first ocatne
\[\int\limits_{0}^{\pi/2}\int\limits_{0}^{2}\int\limits_{0}^{r \cos \theta+4r \sin \theta}rdzdrd \theta\]
you 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
Well that is quite right. I find difficulty in drawing graph in 3d through and then finding it's limit. Any guess.