Question

(a) Find the volume of the given solid.

(b) Express the volume obtained in part (a) in terms of height of the ring.

Short answer

(a) The volume of the given solid is \(\frac{{4\pi }}{3}{\left( {r_2^2 - r_1^2} \right)^{\frac{3}{2}}}\).

(b) The volume of the given solid is \(\frac{\pi }{6}{h^3}\)

Step-by-step solution

Formula used

(a)

If\(f\)is a polar rectangle\(R\)given by\(0 \le a \le r \le b,\alpha \le \theta \le \beta \), where\(0 \le \beta - \alpha \le 2\pi \), then,

If\(g(x)\)is the function of\(x\)and\(h(y)\)is the function of\(y\)then,

\(\int_a^b {\int_c^d g } (x)h(y)dydx = \int_a^b g (x)dx\int_c^d h (y)dy.........(2)\)

Substitute \(x = r\cos \theta \) and \(y = r\sin \theta \) in the equation (1) and obtain the required volume.

Integrate the function with respect to\(r\)and\(\theta \)by using the equation (2).

Let\(t = {r^2}\).

Then,\(dt = 2rdr\).

Obtain the required volume as follows.

\(\begin{aligned}{l}2\int_0^{2\pi } {\int_{{r_1}}^{{r_2}} r } \sqrt {r_2^2 - {r^2}} drd\theta = \frac{2}{2}\int_0^{2\pi } d \theta \int_{{r_1}}^{{r_2}} 2 r\sqrt {r_2^2 - {r^2}} dr\\ = \int_0^{2\pi } d \theta \int_{{r_1}}^{{r_2}} {\sqrt {r_2^2 - t} } dt\\ = (\theta )_0^{2\pi }\left( {\frac{{ - 2{{\left( {r_2^2 - t} \right)}^{\frac{3}{2}}}}}{3}} \right)_{{r_1}}^{{r_2}}\\ = (\theta )_0^{2\pi }\left( {\frac{{ - 2{{\left( {r_2^2 - {r^2}} \right)}^{\frac{3}{2}}}}}{3}} \right)_{{r_1}}^{{r_2}}\end{aligned}\)

Apply the limit of \(r\) and \(\theta \)

\(2\int_0^{2\pi } {\int_{{r_1}}^{{r_2}} r } \sqrt {r_2^2 - {r^2}} drd\theta = (2\pi - 0)\left( {\frac{{ - 2{{\left( {r_2^2 - {{\left( {{r_2}} \right)}^2}} \right)}^{\frac{3}{2}}}}}{3} - \frac{{\left( { - 2{{\left( {r_2^2 - {{\left( {{r_1}} \right)}^2}} \right)}^{\frac{3}{2}}}} \right)}}{3}} \right)\)

\(2\int_0^{2\pi } {\int_{{r_1}}^{{r_2}} r } \sqrt {r_2^2 - {r^2}} drd\theta = (2\pi )\left( {\frac{{ - 2{{(0)}^{\frac{3}{2}}}}}{3} + \frac{{2{{\left( {r_2^2 - r_1^2} \right)}^{\frac{3}{2}}}}}{3}} \right)\)

\(2\int_0^{2\pi } {\int_{{r_1}}^{{r_2}} r } \sqrt {r_2^2 - {r^2}} drd\theta = \frac{{2\pi }}{3}\left( {0 + 2{{\left( {r_2^2 - r_1^2} \right)}^{\frac{3}{2}}}} \right)\)

\(2\int_0^{2\pi } {\int_{{r_1}}^{{r_2}} r } \sqrt {r_2^2 - {r^2}} drd\theta = \frac{{4\pi }}{3}{\left( {r_2^2 - r_1^2} \right)^{\frac{3}{2}}}\)

Thus, the volume of the given solid is \(\frac{{4\pi }}{3}{\left( {r_2^2 - r_1^2} \right)^{\frac{3}{2}}}\).

Formula used(b)

Pythagoras theorem: In a right-angled triangle, square of the hypotenuse is equal to square of the sum of other two sides of the triangle.

Draw the outline sketch of the required region

Use Pythagoras theorem and simplify the equation

\(\begin{aligned}{l}{\left( {\frac{h}{2}} \right)^2} = r_2^2 - r_1^2\\\frac{h}{2} = \sqrt {r_2^2 - r_1^2} \\\frac{{{h^3}}}{8} = {\left( {r_2^2 - r_1^2} \right)^{\frac{3}{2}}}\end{aligned}\)

Hence, the volume becomes,

Thus, the volume of the given solid is \(\frac{\pi }{6}{h^3}\).