Question

Find the volume of the solid enclosed by the surface \(z = x{\sec ^2}y\)and the plane \(x = 0,x = 2,y = 0\& y = \frac{\pi }{4}\)

Short answer

Given:- \(z = x{\sec ^2}y\)

To find:- Volume

Step-by-step solution

Step (1):- Implementing limits in functions

Let V be the volume of the solid. Describe the solid as a range for each of the variable.

\(\begin{array}{l}0 \le x \le 2\\0 \le y \le \frac{\pi }{4}\end{array}\)

Choose z to be the integrand. Set up the integral with either order of integration

V=\(\int\limits_0^{\pi /4} {\int\limits_0^2 {x{{\sec }^2}ydxdy} } \)

Recall that \({\sec ^2}y\)plays the role of a constant multiplier

V=\(\int\limits_0^{\pi /4} {\int\limits_0^2 {x{{\sec }^2}ydxdy} } \)

\(\begin{array}{l}V = \int\limits_0^{\pi /4} {{{\sec }^2}y\left( {\frac{{{x^2}}}{2}} \right)_0^2dy} \\V = \int\limits_0^{\pi /4} {{{\sec }^2}y\left( {2 - 0} \right)dy} \\V = 2\int\limits_0^{\pi /4} {{{\sec }^2}ydy} \end{array}\)

Step (3):- Integrating w.r.t y

\(\begin{array}{l}V = 2\int\limits_0^{\pi /4} {{{\sec }^2}ydy} \\V = 2\int\limits_0^{\pi /4} {\{ (\tan y)dy\} _0^{\pi /4}} \\V = 2\int\limits_0^{\pi /4} {\left( {\tan \frac{\pi }{4} - \tan 0} \right)} \\V = 2\left( {1 - 0} \right) = 2\end{array}\)

Therefore, solids volume is 2 cubic meter.