Question

Evaluate the double integral by first identifying it as the volume of a solid., \(R = \{ x,y| - 2 \le x \le 2,1 \le y \le 6\} \)

Short answer

The volume of a solid is\(60\).

Step-by-step solution

Step-1: - Given data

Given the function is

\(f(x,y) = 3\)

And the rectangular region,

\(R = \{ (x,y)| - 2 \le x \le 2,1 \le y \le 6\} \)

Step-2: - Finding the double integral

\(R = \{ x,y| - 2 \le x \le 2,1 \le y \le 6\} \)

Now,

\(\begin{aligned}\int\limits_{ - 2}^2 {\int\limits_1^6 {3dydx} } &= \int\limits_{ - 2}^2 {\left( {\int\limits_1^6 {3dy} } \right)dxdy} \\ &= \int\limits_{ - 2}^2 {\left( {\left( {3y} \right)_1^6} \right)dx} \\ &= \int\limits_{ - 2}^2 {\left( {3(6) - 3(1)} \right)dx} \\ &= \int\limits_{ - 2}^2 {\left( {18 - 3} \right)dx} \\ &= \int\limits_{ - 2}^2 {15dx} \end{aligned}\)

\(\int\limits_{ - 2}^2 {\int\limits_1^6 {3dydx} } = 15\int\limits_{ - 2}^2 {dx} \)

\(\begin{aligned}{c}\int\limits_{ - 2}^2 {\int\limits_1^6 {3dydx} } &= 15\left( x \right)_{ - 2}^2\\ &= 15(2 + 2)\\ &= 15(4)\\ &= 60\end{aligned}\)

\( \Rightarrow \int\limits_{ - 2}^2 {\int\limits_1^6 {3dydx} } = 60\)

Therefore,\(\) \(\int\limits_{ - 2}^2 {\int\limits_1^6 {3dydx} } = 60\)