Question

Evaluate the double integral by first identifying it as the volume of a solid., \(R = \left\{ {\left( {x,y} \right)/0 \le x \le 5,0 \le y \le 3} \right\}\)

Short answer

Finding the value of integral

,\(R = \left\{ {\left( {x,y} \right)/0 \le x \le 5,0 \le y \le 3} \right\}\)

Step-by-step solution

Given data:

The integral is given by,

, where region is defined as

\(R = \left\{ {\left( {x,y} \right)/0 \le x \le 5,0 \le y \le 3} \right\}\)

The objection is to find the value of integral.

Finding double integral:

Substitute the limits of x and y variable as follows

\(\int\limits_0^3 {\int\limits_0^5 {\left( {5 - x} \right)dxdy = \int\limits_0^3 {\left[ {\int\limits_0^5 {\left( {5 - x} \right)dx} } \right]dy} } } \)

\( = \int\limits_0^3 {\left[ {\left( {5x} \right)_0^5 - \left( {\frac{{{x^2}}}{2}} \right)_0^5} \right]} dy\)

\( = \int\limits_0^3 {\left[ {5\left( {5 - 0} \right) - \frac{1}{2}\left( {{{\left( 5 \right)}^2} - 0} \right)_0^5} \right]} dy\)

\( = \int\limits_0^3 {\left[ {25 - \frac{{25}}{2}} \right]} dy\)

\( = \int\limits_0^3 {\left( {\frac{{50 - 25}}{2}} \right)} dy\)

\( = \int\limits_0^3 {\frac{{25}}{2}} dy\)

\( = \frac{{25}}{2}\int\limits_0^3 {dy} \)

\( = \frac{{25}}{2}\left( y \right)_0^3\)

\(\begin{array}{l} = \frac{{25}}{2}\left( {3 - 0} \right)\\ = \frac{{25}}{2}\left( 3 \right)\end{array}\)

\( \Rightarrow \int\limits_0^3 {\int\limits_0^5 {\left( {5 - x} \right)dxdy = \frac{{75}}{2}} } \)

Hence, the value of integral is

\(\int\limits_0^3 {\int\limits_0^5 {\left( {5 - x} \right)dxdy = \frac{{75}}{2}} } \)