
Evaluate the double integral ∬d(x² + 6y)da, where d is bounded by y = x, y = x³, and x ≥0
Solution:
The area of a closed, bounded plane region R is
A = ∫RdA
The double integral which is of the form aims at finding the area enclosed by curves y = x and y = x³ as shown in the diagram below:
The shaded region is the one for which the area has to be found out. Hence the double integral can be reformulated as below:
Area of the shaded region OAO = ∫10∫y=xy=x3(x² + 6y)dxdy
Solving further we get
= ∫10[x2y+6(y2/2)]y=xy=x3dx
= ∫10[x2(x)+3(x)2−x2(x3)+3(x3)2)]10dx
= ∫10[x3+3x2−x5−3x6]10dx
= [(x4/4)+3(x3)/3−x6/6−3(x7/7)]10
= [14x4+x3−16x6−37x7]10dx
= [(1 / 4) + 1 - (1 / 6) - (3 / 7) - 0]
= [ (5 / 4) - (1 / 6) - (3 / 7)]
= 105−14−3684
= 55/84
Evaluate the double integral ∬d(x² + 6y)da, where d is bounded by y = x, y = x³, and x ≥0
Summary:
Solving the double integral ∬d(x² + 6y)da, where d is bounded by y = x, y = x³, and x ≥0 as the solution obtained is 55/84
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