Question

Evaluate the double integral \(\iint\limits_D {\left( {x{y^2}} \right)dA}\)D is enclosed by\(x = 0, x = \sqrt {1 - {y^2}} \)

Short answer

The solution of the given integral can be:

\(\iint\limits_D {x{y^2}dA} = \frac{2}{{15}}\)

Step-by-step solution

Step 1:Find the limits

Consider the integral

\(\iint\limits_D {x{y^2}dA},x = 0,x = \sqrt {1 - {y^2}} \)

Find the values of y equate \(x = 0\).

\(\begin{array}{l}x = \sqrt {1 - {y^2}} \\0 = \sqrt {1 - {y^2}} \\1 - {y^2} = 0\\y = \pm 1\end{array}\)

Set up x as a function of y.

Check for the end points, the upper and lower bounds of x and y respectively as:

\(\begin{array}{l}0 \le x \le \sqrt {1 - {y^2}} \\ - 1 \le y \le 1\end{array}\)

Chose the ordering of x’s limits because for \( - 1 \le y \le 1,0 \le x \le \sqrt {1 - {y^2}} \)

Rewrite the double integral as:

\(\iint\limits_D {x{y^2}dA} = \int\limits_{ - 1}^1 {\int\limits_0^{\sqrt {1 - {y^2}} } {x{y^2}} } dxdy\)

Evaluate the integral

\(\iint\limits_D {x{y^2}dA} = \int\limits_{ - 1}^1 {\int\limits_0^{\sqrt {1 - {y^2}} } {x{y^2}} } dxdy\)

\(\iint\limits_D {x{y^2}dA} = \int\limits_{ - 1}^1 {\left[ {\frac{{{x^2}{y^2}}}{2}} \right]_0^{\sqrt {1 - {y^2}} }} dy\)

\(\iint\limits_D {x{y^2}dA} = \int\limits_{ - 1}^1 {\left[ {\frac{{{{\left( {\sqrt {1 - {y^2}} } \right)}^2}{y^2}}}{2} - \frac{{0{y^2}}}{2}} \right]} dy\)

\(\iint\limits_D {x{y^2}dA} = \int\limits_{ - 1}^1 {\frac{{(1 - {y^2}){y^2}}}{2}} dy\)

\(\iint\limits_D {x{y^2}dA} = \int\limits_{ - 1}^1 {\frac{{{y^2} - {y^4}}}{2}} dy\)

\(\iint\limits_D {x{y^2}dA} = \frac{1}{2}\left[ {\frac{{{y^3}}}{3} - \frac{{{y^5}}}{5}} \right]_{ - 1}^1\)

\(\iint\limits_D {x{y^2}dA} = \frac{1}{2}\left[ {\left( {\frac{{{1^3}}}{3} - \frac{{{1^5}}}{5}} \right) - \left( {\frac{{ - {1^3}}}{3} - \frac{{ - {1^5}}}{5}} \right)} \right]\)

\(\iint\limits_D {x{y^2}dA} = \frac{1}{2}\left( {\frac{1}{3} - \frac{1}{5} + \frac{1}{3} - \frac{1}{5}} \right)\)

\(\iint\limits_D {x{y^2}dA} = \frac{1}{2}\left( {\frac{2}{3} - \frac{2}{5}} \right)\)

\(\iint\limits_D {x{y^2}dA} = \frac{2}{{15}}\)

Hence the required value is:.

\(\iint\limits_D {x{y^2}dA} = \frac{2}{{15}}\)