Question

(a) To set up: an integral function for the volume of the solid obtained by rotating the region bounded by the given curve.

(b)To evaluate: The integral function.

Short answer

(a). The volume of the solid obtained by rotation of the region bounded by the curve is \(V = 2\pi \int_{ - 3}^3 {(5 - y)} \left( {4 - \sqrt {7 + {y^2}} } \right)dy\).

(b) The value of the integral is \(163.02712\).

Step-by-step solution

Given information

(a) We have been given the equation \({x^2} - {y^2} = 7 \Rightarrow x = \pm \sqrt {7 + {y^2}} \).

(b) From part (a) the integral of the volume \(V\)is given by \(V = 2\pi \int_{ - 3}^3 {(5 - y)} \left( {4 - \sqrt {7 + {y^2}} } \right)dy\).

The concept of volume by cylindrical shell method

Definition of Volume by cylindrical shell method:

Consider that the solid \(S\) is obtained by rotation of the region under the curve \(y = f(x)\) about the \(y\)-axis from \(a\) to \(b\) is \(V = \int_a^b 2 \pi xf(x)dx\).

Draw the graph

(a)

We know that \({x^2} - {y^2} = 7 \Rightarrow x = \pm \sqrt {7 + {y^2}} \)

We are interested only in the right lobe, 50 only \(x = \sqrt {7 + {y^2}} \) is relevant

The shaded region in the graph is bounded by \(x = \sqrt {7 + {y^2}} \) and \(x = 4\)

We have to rotate the black shaded region about the red dashed line. \((y = 5)\)

Find the expression for volume

(a)

When we rotate the blue horizontal strip about \(y = 5\) (red dashed line) (as shown in figure), we get a cylindrical shell with radius\( = 5 - y\) And height \( = 4 - \sqrt {7 + {y^2}} \).

Notice that: Radius = Distance between red and blue lines Height = Length of the blue horizontal strip. Height = Length of the blue horizontal strip, which meets \(x = \sqrt {7 + {y^2}} \) on the left and \(x = 4\) on the right.

\(V = 2\pi \int_0^{\pi /4} {{\rm{ (Radius) (Height) }}} dx\)

Integrate with respect to \(y\) because the shells are parallel to the \(y\)-axis. Integrate from \( - 3\) to 3 because the shaded region exists for \(y \in ( - 3,3)\)

\(V = 2\pi \int_{ - 3}^3 {(5 - y)} \left( {4 - \sqrt {7 + {y^2}} } \right)dy\)

find the value of integral function

(b)

Rearrange Equation.

\(\begin{aligned}{}V = 2\pi \int_{ - 3}^3 {(y - 5)} \left( {\sqrt {{y^2} + 7} - 4} \right)dy\\ = 2\pi \int_{ - 3}^3 5 \left( {4 - \sqrt {{y^2} + 7} } \right) - y\left( {4 - \sqrt {{y^2} + 7} } \right)dy\\ = 2\pi \left( {\int_{ - 3}^3 5 \left( {4 - \sqrt {{y^2} + 7} } \right)dy - \int_{ - 3}^3 y \left( {4 - \sqrt {{y^2} + 7} } \right)dy} \right)\end{aligned}\) ……(1)

Integrate Equation (1)

\(\begin{aligned}{}V = 2\pi \left( {\begin{aligned}{{}{}}{5\left( {4y - 7\left( {\frac{1}{{14}}y\sqrt {7 + {y^2}} + \frac{1}{2}\ln \left( {\frac{1}{{\sqrt 7 }}y + \sqrt {1 + \frac{1}{7}{y^2}} } \right)} \right)} \right.}\\{\frac{1}{2}\left( {4\left( {{y^2} + 7} \right) - \frac{2}{3}{{\left( {{y^2} + 7} \right)}^{\frac{3}{2}}}} \right)}\end{aligned}} \right)_{ - 3}^3\\ = \left( {\begin{aligned}{{}{}}{\left. {40\pi y - 5\pi y\sqrt {7 + {y^2}} - 35\pi \ln \left( {\frac{1}{{\sqrt 7 }}y + \sqrt {1 + \frac{1}{7}{y^2}} } \right)} \right)_{ - 3}^3}\\{ - 4\pi {y^2} - 28\pi + \pi \frac{2}{3}{{\left( {{y^2} + 7} \right)}^{\frac{3}{2}}}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{{}{}}{\left( {120\pi - (60\pi ) - \left( {\frac{{35}}{2}\pi \ln (7)} \right) - 36\pi - 28\pi + \frac{{128\pi }}{3}} \right)}\\{ - \left( { - 120\pi - ( - 60\pi ) - \left( { - \frac{{35}}{2}\pi \ln (7)} \right) - 36\pi - 28\pi + \frac{{128\pi }}{3}} \right)}\end{aligned}} \right)\\ = \left( {\frac{{116}}{6} - \frac{{35}}{2}\pi \ln (7)} \right) - \left( {\frac{{35}}{2}\pi \ln (7) - 244} \right)\end{aligned}\)

Simplify further,

\(\begin{aligned}{}V = \frac{\pi }{6}(232 - 105\ln (7)) - \frac{\pi }{6}(105\ln (7) - 488)\\ = 5\pi (24 - 7\ln (7))\\ = 163.02712\end{aligned}\)

The value of the integral is \(163.02712\).