Question

Evaluate the double integral \(\iint\limits_D {\left( {2x - y} \right)dA}\)D is bounded by the circle with the centre origin and radius 2.

Short answer

The solution of the given integral can be:

\(\iint\limits_D {\left( {2x - y} \right)dA} = 0\)

Step-by-step solution

Find the limits

Consider the integral

\(\iint\limits_D {\left( {2x - y} \right)dA}\)

D is bounded by the circle with the centre origin and radius 2.

So we can write,

\(\begin{array}{l}{\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} = {2^2}\\{x^2} + {y^2} = 4\\x = \pm \sqrt {4 - {y^2}} \end{array}\)

Sketch of the region bounded by the circle with centre the origin and radius 2 is shown below:

The region D can be expressed as follows:

\(D = \{ (x,y)| - 2 \le y \le 2, - \sqrt {4 - {y^2}} \le x \le \sqrt {4 - {y^2}} \} \)

Rewrite the double integral as:

\(\iint\limits_D {\left( {2x - y} \right)dA} = \int\limits_{ - 2}^2 {\int\limits_{ - \sqrt {4 - {y^2}} }^{\sqrt {4 - {y^2}} } {\left( {2x - y} \right)} } dxdy\)

Step 2:Evaluate the integral

\(\iint\limits_D {\left( {2x - y} \right)dA} = \int\limits_{ - 2}^2 {\int\limits_{ - \sqrt {4 - {y^2}} }^{\sqrt {4 - {y^2}} } {\left( {2x - y} \right)} } dxdy\)

\(\iint\limits_D {\left( {2x - y} \right)dA} = \int\limits_{ - 2}^2 {\left[ {{x^2} - xy} \right]_{ - \sqrt {4 - {y^2}} }^{\sqrt {4 - {y^2}} }} dy\)

\(\iint\limits_D {\left( {2x - y} \right)dA} = \int\limits_{ - 2}^2 {\left[ {\left( {4 - {y^2} - y\sqrt {4 - {y^2}} } \right) - \left( {4 - {y^2} + y\sqrt {4 - {y^2}} } \right)} \right]} dy\)

\(\iint\limits_D {\left( {2x - y} \right)dA} = \int\limits_{ - 2}^2 {\left[ {\left( {4 - {y^2}} \right) - y\sqrt {4 - {y^2}} - \left( {4 - {y^2}} \right) - y\sqrt {4 - {y^2}} } \right]} dy\)

\(\iint\limits_D {\left( {2x - y} \right)dA} = \int\limits_{ - 2}^2 {\left[ { - 2y\sqrt {4 - {y^2}} } \right]} dy\)

\(\iint\limits_D {\left( {2x - y} \right)dA} = \frac{2}{3}\left[ {{{\left( {4 - {y^2}} \right)}^{\frac{3}{2}}}} \right]_{ - 2}^2\)

\(\iint\limits_D {\left( {2x - y} \right)dA} = \frac{2}{3}\left[ {{{\left( {4 - 4} \right)}^{\frac{3}{2}}} - {{\left( {4 - 4} \right)}^{\frac{3}{2}}}} \right]_{ - 2}^2\)

\(\iint\limits_D {\left( {2x - y} \right)dA} = 0\)

Hence the required value is:\(\iint\limits_D {\left( {2x - y} \right)dA} = 0\)