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Q16E

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Essential Calculus: Early Transcendentals
Found in: Page 707
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

16: Evaluate the double integral \(\iint\limits_D {\left( {{x^2} + 2y} \right)dA}\) D is bounded by\(y = x,y = {x^3},x \ge 0\)

The solution of the given integral can be: \(\iint\limits_D {\left( {{x^2} + 2y} \right)dA} = \frac{{23}}{{84}}\)

See the step by step solution

Step by Step Solution

Step 1:Find the limits

Consider the integral

\(\iint\limits_D {\left( {{x^2} + 2y} \right)dA},y = x,y = {x^3}x \geqslant 0\)

The region D is shown in the figure.

Evaluate the limit of x as shown below:

\(\begin{array}{l}{x^3} = x\\{x^3} - x = 0\\\left( {{x^2} - 1} \right)x = 0\\x = 0\,,x = 1\end{array}\)

Since \(x \ge 0\)

The region D can be expressed as:\(D = \{ (x,y)|0 \le x \le 1,{x^3} \le y \le x\} \)

Step 2:Evaluate the integral

\(\begin{aligned}\iint\limits_D {\left( {{x^2} + 2y} \right)dA} &= \int\limits_0^1 {\int\limits_{{x^3}}^x {\left( {{x^2} + 2y} \right)dydx} } \hfill \\\iint\limits_D {\left( {{x^2} + 2y} \right)dA} &= \int\limits_0^1 {\left( {{x^2}y + {y^2}} \right)_{{x^3}}^xdx} \hfill \\\iint\limits_D {\left( {{x^2} + 2y} \right)dA} &= \int\limits_0^1 {\left( {{x^2}(x) + {{(x)}^2} - ({x^2}({x^3}) + {{({x^3})}^2})} \right)dx} \hfill \\\iint\limits_D {\left( {{x^2} + 2y} \right)dA} &= \int\limits_0^1 {\left( {{x^3} + {x^2} - {x^5} - {x^6}} \right)dx} \hfill \\ \iint\limits_D {\left( {{x^2} + 2y} \right)dA} &= \left( {\frac{1}{4}{x^4} + \frac{1}{3}{x^3} - \frac{{{x^6}}}{6} -\frac{{{x^7}}}{7}} \right)_0^1 \hfill \\\iint\limits_D {\left( {{x^2} + 2y} \right)dA} &= \frac{1}{4} + \frac{1}{3} - \frac{1}{6} - \frac{1}{7} \hfill \\\iint\limits_D {\left( {{x^2} + 2y} \right)dA} &= \frac{{23}}{{84}} \hfill \\ \end{aligned} \)

Hence the required value is: \(\iint\limits_D {\left( {{x^2} + 2y} \right)dA} = \frac{{23}}{{84}}\)

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