Find the volume of the region in the first octant bounded below by the xy - plan

Question

Find the volume of the region in the first octant bounded below by the xy - plane , on the sides by x = 0 and y = 2x and above by y2 + z2 = 16.

Explanation / Answer

volume => triple integral dzdydx now in first octant x,y,z all are positive so we get the limit of z = 0 to sqrt(16-y^2) and y= 0 to 2x and x=0 to 2 so integral value is sqrt(16-y^2) dydx => 8sin^(-1)(y/4) + ysqrt(16-y^2)/2 dx putting limits of y we get => 8sin^(-1)(x/2) + xsqrt(16-4x^2)dx so we the value of integral as 8xsin^(-1)(x/2) +8sqrt(1-(1/4)x^2) + xsqrt(16-4x^2)^(2/3)/12 putting limits 8pi + 8 thus 8(pi +1) is the ansqwer

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