Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

9. \({x^2} - {y^2} = {a^2},x = a + h\)\((wherea > 0,h > 0);\)about the \( y - \)axis

Short answer

The equation of the curves is \(V = \frac{4}{3}\pi {\left( {2ah + {h^2}} \right)^{3/2}}\)

Step-by-step solution

To solve the intersection

The equation is,

\({x^2} - {y^2} = {a^2},x = a + h,a > 0,h > 0,\)Rotation about \( y - \)axis

Solve for \(y\)

\(y = \pm \sqrt {{x^2} - {a^2}} \)

We set the equations for \(y\)equal to each other to solve for intersections

\(\sqrt {{x^2} - {a^2}} = - \sqrt {{x^2} - {a^2}} \)

This is one limit (bound) for the integral. Note that even though you get \( \pm a\)as intersections, \(a > 0\)was stated in the question.

\(x = a\)

Formula to solve volume rotating about y-axis and when you solve for\(y\). \(a\)and \(a + h\)are our limits.

\(V = 2\pi \int_a^{a + h} x {y_2}dx - 2\pi \int_a^{a + h} x {y_1}dx\)

Plug in \(\sqrt {{x^2} - {a^2}} \) for \({y_2}\)and \( - \sqrt {{x^2} - {a^2}} \) for \({y_1}\)

\(V = 2\pi \int_a^{a + h} x \sqrt {{x^2} - {a^2}} dx - 2\pi \int_a^{a + h} x \left( { - \sqrt {{x^2} - {a^2}} } \right)dx\)

Simplify the Sign

Simplify the - sign

\(V = 2\pi \int_a^{a + h} x \sqrt {{x^2} - {a^2}} dx + 2\pi \int_a^{a + h} x \left( {\sqrt {{x^2} - {a^2}} } \right)dx\)

Left and right are the same integral, condense into\( 2\)times that integral.

\(V = 4\pi \int_a^{a + h} x \sqrt {{x^2} - {a^2}} dx\)

We're using substitution here, just with a different letter.

Set\(z = {x^2} - {a^2}\)and\(dz = 2xdx\)

We need to solve for limits when doing substitution

Plug in\(a\)for and\(x\)\(a + h\)for\( x.\)

\({z_1} = 0,{z_2} = 2ah + {h^2}\)

\(V = 2\pi \int_0^{2ah + {h^2}} {\sqrt z } dz\)

Final proof

Solving the integral

\( = \left. {\frac{4}{3}\pi {z^{3/2}}} \right|_0^{2ah + {h^2}}\)

Plug in the limits,

\(v = \frac{4}{3}\pi {\left( {2ah + {h^2}} \right)^{3/2}}\)