Question

\(\int {\int\limits_D y dA} \).

Short answer

The process of finding a function \(g\left( x \right)\)the deviate of which, \(Dg\left( x \right)\)is equal to a given function\(f\left( x \right)\).

Step-by-step solution

Given data:

Consider,

\(x = y - {y^3},y = {\left( {x + 1} \right)^2}\)

The region of integration\(D\)can be partitioned into the union of two regions graph:

\(\begin{array}{l}I = \int {\int\limits_D {ydA} } \\ = \int {\int\limits_{{D_1}} {ydA} } + \int {\int\limits_{{D_2}} {ydA} } \end{array}\)

Both Region should be integrated first on\(x\). For\({D_1}\)the equation for the left boundary needs to be inverted given\(\sqrt y - 1 \le x \le y - {y^3}\)and for\({D_2}\)the boundaries are\( - 1 \le x \le y - {y^3}\). Set up the limits of the double integral.

\(I = \int\limits_0^1 y \int\limits_{\sqrt y - 1}^{y - {y^3}} {dxdy} + \int\limits_{ - 1}^0 y \int\limits_{ - 1}^{y - {y^3}} {dy} \)

Integrate on\('x'\)

\(\begin{array}{l}I = \int\limits_\infty ^1 {y\left( {y - {y^3} - \sqrt y + 1} \right)} dy + \int\limits_{ - 1}^0 {y\left( {y - {y^3} + 1} \right)} dy\\ = \int\limits_\infty ^1 {\left( { - y + {y^2} - {y^{{3 \mathord{\left/{\vphantom {3 2}} \right.} 2}}} + y} \right)} dy + \int\limits_{ - 1}^0 {\left( { - {y^4} + {y^2} + y} \right)dy} \end{array}\)

Integrate on\('y'\)

\(\begin{array}{l}I = \left( {\frac{{ - 1}}{5}{y^5} + \frac{1}{3}{y^3} - \frac{2}{5}{y^{{5 \mathord{\left/{\vphantom {5 2}} \right.} 2}}} + \frac{1}{2}{y^2}} \right)_0^1 + \left( {\frac{{ - 1}}{5}{y^5} + \frac{1}{3}{y^3} + \frac{1}{2}{y^2}} \right)_{ - 1}^0\\ = \frac{{ - 1}}{5} + \frac{1}{3} - \frac{2}{5} + \frac{1}{2} - 0 + 0 - \frac{1}{5} + \frac{1}{3} - \frac{1}{2}\\ = - \frac{1}{5} + \frac{1}{3} - \frac{2}{5} + \frac{1}{2} - \frac{1}{5} + \frac{1}{3} - \frac{1}{2}\\ = - \frac{2}{{15}}\end{array}\)

Hence, the value of integral is\( - \frac{2}{{15}}\).