Question

Determine the solid whose volume is given by the iterated integral and evaluate it.

Short answer

The value of the given iterated integral is \(\frac{\pi }{3}\).

Step-by-step solution

Given data

The region \(D\) is \(\left\{ {(\rho ,f,\theta )\mid 0 \le \rho \le \sec f,0 \le f \le \frac{\pi }{4},0 \le \theta \le 2\pi } \right\}\).

Concept used

If \(g(x)\) is the function of \(x\) and \(h(y)\) is the function of \(y\) and \(k(z)\) is the function of \(z\) then, \(\int_a^b {\int_c^d {\int_e^f g } } (x)h(y)k(z)dzdydx = \int_a^b g (x)dx\int_c^d h (y)dy\int_e^f k (z)dz\) ……. (1)

Step 3:Sketch of the given region \(D\)

From the given condition as follows:

\(\begin{aligned}\rho = \sec \Phi \\\frac{\rho }{{\sec \Phi }} = 1\\\rho \cos \Phi = 1\\z = 1\end{aligned}\)

Since \(\theta \) varies between 0 and \(2\pi \) and \(f = \frac{\pi }{4}\), the region lies under the plane \(z = 1\) and inside the cone \(\frac{\pi }{4}\).

So, the outline sketch of the given region \(D\) is shown in the figure 1 as follows:

Figure 1

Find the equation of \(\int_0^{\frac{\pi }{4}} {\int_0^{2\pi } {\int_0^{\sec \Phi } {{\rho ^2}} } } \sin \Phi d\rho d\theta d\Phi \)

Use the equation (1) and separate the given iterated integral and then integrate with respect to \(\theta \) and \(\rho \) as follows:

\(\begin{aligned}\int_0^{\frac{\pi }{4}} {\int_0^{2\pi } {\int_0^{\sec f} {{\rho ^2}} } } \sin \Phi \;d\rho \;d\theta \;d\Phi &= \int_0^{2\pi } d \theta \int_0^{\frac{\pi }{4}} {\int_0^{\sec f} {{\rho ^2}} } \sin \Phi \;d\rho \;d\Phi \\ &= (\theta )_0^{2\pi }\int_0^{\frac{\pi }{4}} {\left( {\frac{{{\rho ^3}\sin \Phi }}{3}} \right)_0^{\sec f}} d\Phi \\ &= (2\pi - 0)\int_0^{\frac{\pi }{4}} {\left( {\frac{{{{\sec }^3}\Phi \sin \Phi }}{3} - \frac{{{{(0)}^3}\sin \Phi }}{3}} \right)} d\Phi \\ &= (2\pi )\int_0^{\frac{\pi }{4}} {\left( {\frac{{{{\sec }^2}\Phi \tan \Phi }}{3} - 0} \right)} d\Phi \end{aligned}\)

Simplify further, it obtains, \(\int_0^{\frac{\pi }{4}} {\int_0^{2\pi } {\int_0^{2\sec f} {{\rho ^2}} } } \sin fd\rho d\theta df = \frac{{2\pi }}{3}\int_0^{\frac{\pi }{4}} {{{\sec }^2}} f\tan fdf\).

Find the value of \(\int_0^{\frac{\pi }{4}} {\int_0^{2\pi } {\int_0^{\sec \Phi } {{\rho ^2}} } } \sin \Phi d\rho d\theta d\Phi \)

Substitute \(t = \tan f,dt = {\sec ^2}fdf\) and integrate with respect to \(f\) as follows:

\(\begin{aligned}\int_0^{\frac{\pi }{4}} {\int_0^{2\pi } {\int_0^{\sec \Phi } {{\rho ^2}} } } \sin \Phi \;d\rho \;d\theta \;d\Phi &= \frac{{2\pi }}{3}\int_0^{\frac{\pi }{4}} t dt\\ &= \frac{{2\pi }}{3}\left( {\frac{{{t^2}}}{2}} \right)_0^{\frac{\pi }{4}}\\ &= \frac{{2\pi }}{3}\left( {\frac{{{{\tan }^2}\Phi }}{2}} \right)_0^{\frac{\pi }{4}}\end{aligned}\)

Find the value of the given iterated integral 

Apply the limit value of \(f\) as follows:

\(\begin{aligned}\int_0^{\frac{\pi }{4}} {\int_0^{2\pi } {\int_0^{\sec f} {{\rho ^2}} } } \sin f\;d\rho \;d\theta \;df &= \frac{{2\pi }}{{3(2)}}\left( {{{\tan }^2}\left( {\frac{\pi }{4}} \right) - \tan (0)} \right)\\ &= \frac{\pi }{3}(1 - 0)\\ &= \frac{\pi }{3}(1)\\ &= \frac{\pi }{3}\end{aligned}\)

Thus, the value of the given iterated integral is \(\frac{\pi }{3}\).