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 = $\int_{R}^{}$dA

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 = $\int_{0}^{1}\int_{y=x^{3}}^{y=x}$(x² + 6y)dxdy

Solving further we get

= $\int_{0}^{1}[x^{2}y + 6(y^{2}/2)]_{y=x^{3}}^{y=x}dx$

= $\int_{0}^{1}[x^{2}(x)+3(x)^{2} - x^{2}(x^{3}) + 3(x^{3})^{2})]_{0}^{1}dx$

= $\int_{0}^{1}[x^{3}+3x^{2} - x^{5} - 3x^{6}]_{0}^{1}dx$

= $[(x^{4}/4) + 3(x^{3})/3 - x^{6}/6 - 3(x^{7}/7)]_{0}^{1}$

= $[\frac{1}{4}x^{4} +x^{3} - \frac{1}{6}x^{6} - \frac{3}{7}x^{7}]_{0}^{1}dx$

= [(1 / 4) + 1 - (1 / 6) - (3 / 7) - 0]

= [ (5 / 4) - (1 / 6) - (3 / 7)]

= $\frac{105 - 14 - 36}{84}$

= 55/84

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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