# Find the volume of the solid generated by revolving the shaded region about the y-axis.

This article aims to find the volume of the solid formed by rotating the shaded region about the y-axis. The article uses the concept of volume of the solid. The volume of the solid generated by a region under $f(x)$ bounded by the y-axis and vertical lines $y=a$ and $y=b$, which is revolved about the y-axis, is:

$V = \int A dx$

Where:

$A = \pi r ^ { 2 } \: and \: r = f(x)$

$V = \pi \int_{ a } ^ { b } x ^ { 2 } dy$

Volume

Area of surface

The given curve is:

$y = 1, x= 0 , x = 4 \tan(\dfrac { \pi } { 3 } ) y$

Find the volume of solid formed by rotating the shaded region about the y-axis.

$V = \int_{ 0 } ^ { 1 } \pi (4 \tan(\dfrac{\pi}{3})y) ^ { 2 } dy$

$= 16 \int_{0}^{1} \tan ^ { 2 } (\dfrac{ \pi } { 3 } y) dy$

Let:

$\dfrac{\pi}{3}y = z , \dfrac{\pi}{3}dy \Rightarrow = dz$

$y=0 \Rightarrow z= 0\: and \: y =1 \Rightarrow z = \dfrac{\pi}{3}$

$V = 16\pi \int_{0} ^ { \dfrac { \pi } { 3 } } \tan ^ { 2 } z ( \dfrac { 3 }{ \pi } ) dz = 48 \int_{ 0 } ^ { \ dfrac { \pi } { 3 } } \tan ^ { 2 } z \: dz$

Since:

$\sec ^ { 2 } x – \tan ^ { 2 } x = 1$

$=48 \int_{0} ^ { \dfrac { \pi}{3}} \sec^{2} z \: dz \:- 48\: \int_{0}^{\dfrac{\pi}{3}} 1 \:dz$

$= 48 \tan z | _{ 0 } ^{ \dfrac { \pi } { 3 } } – \: 48 z |_{0} ^ { \dfrac { \pi }{3}}$

$= 48 ( \tan (\dfrac{ \pi } { 3 }) – \tan 0) – \:48(\dfrac{ \pi }{ 3 } – 0)$

$= 48 (\sqrt { 3 } -0) – 48 \dfrac{ \pi } { 3 }$

$= 48(\sqrt { 3 } – \dfrac{ \pi } { 3 })$

The volume of solid generated by revolving the shaded region is $48(\sqrt {3} – \dfrac{\pi}{3})$.

## Numerical Result

The volume of solid generated by revolving the shaded region is $48(\sqrt {3} – \dfrac{\pi}{3})$.

## Example

Find the volume of solid generated by revolving the shaded region about the y-axis.

Solution

The given curve is:

$y = 1, x= 0 , x = 5 \tan(\dfrac{\pi}{3})y$

Find the volume of solid formed by rotating the shaded region about the y-axis.

$V = \int_{0}^{1} \pi (5 \tan(\dfrac{\pi}{3})y)^{2} dy$

$= 25 \int_{0}^{1} \tan^{2} (\dfrac{\pi}{3} y) dy$

Let:

$\dfrac{\pi}{3}y = z , \dfrac{\pi}{3}dy \Rightarrow = dz$

$y=0 \Rightarrow z= 0\: and \: y =1 \Rightarrow z = \dfrac{\pi}{3}$

$V = 25\pi \int_{0}^{\dfrac{\pi}{3}} \tan ^{2} z (\dfrac{3}{\pi})dz = 75 \int_{0}^{\dfrac{\pi}{3}} \tan^{2} z \: dz$

Since:

$\sec ^{2} x – \tan ^{2} x = 1$

$=75 \int_{0}^{\dfrac{\pi}{3}} \sec^{2} z \: dz \:- 75\: \int_{0}^{\dfrac{\pi}{3}} 1 \:dz$

$= 75 \tan z | _{0}^{\dfrac{\pi}{3}} – \: 75 z |_{0}^{\dfrac{\pi}{3}}$

$= 75 (\tan (\dfrac{\pi}{3}) – \tan 0) – \:75 (\dfrac{\pi}{3} – 0)$

$= 75 (\sqrt {3} -0) – 75 \dfrac{\pi}{3}$

$= 75(\sqrt {3} – \dfrac{\pi}{3})$

The volume of solid generated by revolving the shaded region is $75(\sqrt {3} – \dfrac{\pi}{3})$.