Evaluate the double integral by first identifying it as the volume of a solid.

Question

∫∫ 2x+1 dA 

 

R= { (x,y): 0≤x≤2 , 0≤y≤1 } 

Answer 1

The double integral will yield the volume under the plane z=2x+1 bounded by the lines x=0 and x=2 and y=0 and y=2 in the xy plane.  

 

Since we are integrating over a rectangular area, the integral simply becomes ∫¹₀∫₀^{2 (}2x+1)dxdy.  The inner integral ∫₀²(2x+1)dx which evaluates to 6.  Now the outer integral becomes ∫₀¹ 6dy which evaluates to 6.

 

The volume under the plane z=2x+1 over the rectangle bounded by x=0 and x=2 and y=0 and y=1 is 6.