Question

Find the volume of the solid enclosed by the surface \(z = 1 + {e^x}\sin y\) and the planes \(x = \pm 1,y = 0,y = \pi \& z = 0\)

Short answer

Given:- \(z = 1 + {e^x}\sin y\)

\(x = \pm 1,y = 0,y = \pi \& z = 0\)

To find:- Volume

Step-by-step solution

Step (1):- Determining the Volume

\(\begin{array}{l}V = \int\limits_{ - 1}^1 {\int\limits_0^1 {\left( {1 + {e^x}\sin y} \right)dydx} } \\V = \int\limits_{ - 1}^1 {\left( {\int\limits_0^1 {\left( {1 + {e^x}\sin y} \right)dy} } \right)dx} \\V = \int\limits_{ - 1}^1 {\left( {y - {e^x}\cos y} \right)_0^\pi dx} \\V = \int\limits_{ - 1}^1 {\left( {\left( {\pi - {e^x}} \right) - \left( {0 - {e^x}} \right)} \right)dx} \\V = \int\limits_{ - 1}^1 {\left( {\pi + 2{e^x}} \right)dx} \\V = \left( {\pi x + 2{e^x}} \right)_{ - 1}^1\\V = \left( {\pi - 2e} \right) - \left( { - \pi + 2{e^{ - 1}}} \right)\\V = 2\pi + 2e - \frac{2}{e}\end{array}\)

Therefore, the volume of the solid is \(2\pi + 2e - \frac{2}{e}\) cubic units.