Evaluate the double integral ∫∫D(x2+y2)dA, where D is bounded by y=x, y=x3, and x≥0

In the region D, x varries from 0 to 1 and y varries from y=x3 to x=y ∴∫∫D(x2+y2)dA=∫x=01∫y=x3x(x2+2y)dydx =∫x=01[x2y+y2]{x3}xdx =∫x=01[x2(x−x3)+[x2−(x3)2]]dx=∫x=01[x3−x5+x2−x6]dx x44−x66+x33−x77∣01=14−16+13−17=2384=0.27381

Best solutions to - "Evaluate the double integral ∫∫D(x2+y2)dA, where D is..."

diferira7c

Answered question

2021-12-16

Evaluate the double integral

\int \int_{D} (x^{2} + y^{2}) d A

, where D is bounded by

y = x, y = x^{3},

and

x \geq 0

Answer & Explanation

Vivian Soares

Beginner2021-12-17Added 36 answers

In the region D, x varries from 0 to 1 and y varries from

y = x^{3}

x = y

∴ \int \int_{D} (x^{2} + y^{2}) d A = \int_{x = 0}^{1} \int_{y = x^{3}}^{x} (x^{2} + 2 y) d y d x

= \int_{x = 0}^{1} {[x^{2} y + y^{2}]}_{{x^{3}}}^{x} d x

= \int_{x = 0}^{1} [x^{2} (x - x^{3}) + [x^{2} - {(x^{3})}^{2}]] d x = \int_{x = 0}^{1} [x^{3} - x^{5} + x^{2} - x^{6}] d x

\frac{x^{4}}{4} - \frac{x^{6}}{6} + \frac{x^{3}}{3} - \frac{x^{7}}{7} ∣_{0}^{1} = \frac{1}{4} - \frac{1}{6} + \frac{1}{3} - \frac{1}{7} = \frac{23}{84} = 0.27381

autormtak0w

Beginner2021-12-18Added 31 answers

Why is the bound from

x^{3}

to x and not vice versa?

nick1337

Expert2021-12-28Added 777 answers

Because They evaluated the integral Vertically which the bounds would the graphs, top graph over bottom graph. Which would be x over $x^{3}$ . x being upper bound, $x^{3}$ being lower bound. Hope this helps.

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