Answer to Question #16165 in Integral Calculus for diey
∫∫∫2y2z dxdzdy ?
∫∫∫dzdxdy ?
1
2012-10-10T08:53:37-0400
∫∫∫2y2z dxdzdy =∫∫(∫2y2z dx)dzdy=∫∫ (4yz*x+c1) dzdy=∫ (∫ (4yx*z+c1) dz) dy=∫
(2yx*z^2+c1*z+c2)dy=∫
([2xz^2]*y+[c1*z+c2])dy=xz^2*y^2+[c1*z+c2]*y+c3=xz^2y^2+c1*z*y+c2*y+c3
∫∫∫dzdxdy=∫∫(∫dz)dxdy=∫∫(z+c)dxdy=∫((z+c)x+c2)dy=((z+c)x+c2)*y+c3
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