Question

To find: The Volume of the solid which is obtained on rotating the region bounded by the given curves about \(x\)-axis.

Short answer

The volume of the solid is \(\frac{{27}}{2}\pi \).

Step-by-step solution

Given information

The function is \(x + y = 3,x = 4 - {(y - 1)^2}\).

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\(V = \int_a^b 2 \pi xf(x)dx\).

Calculate the volume

It is given that the curves that bound a region are \(x + y = 3,x = 4 - {(y - 1)^2}\).

The specific line around which the region is rotated is\(x\)-axis.

Equate the values of\(x\)and obtain the value of\(y\):

\(\begin{aligned}3 - y &= 4 - {(y - 1)^2}\\{(y - 1)^2} - y - 1 &= 0\\{y^2} - 2y + 1 - y - 1 &= 0\\{y^2} - 3y &= 0\end{aligned}\)

Then,\(y = 0,y = 3\).

Here, the shell radius is\(y\)and the shell height is

\(\begin{aligned}4 &- {(y - 1)^2} - (3 - y)\\4 &- {y^2} + 2y - 1 - 3 + y\\3y &- {y^2}\end{aligned}\)

The region lies between\(y = 0\), and\(y = 3\), so by definition of cylindrical shell, the volume becomes

\(\begin{aligned}V &= \int_0^3 2 \pi y\left( {3y - {y^2}} \right)dy\\ &= 2\pi \int_0^3 {\left( {3y - {y^2}} \right)} dy\\ &= 2\pi \left[ {{y^3} - \frac{{{y^4}}}{4}} \right]_0^3\\ &= 2\pi \left( {27 - \frac{{81}}{4}} \right)\end{aligned}\)

On further simplifications,

\(\begin{aligned}V &= 2\pi \left( {\frac{{108 - 81}}{4}} \right)\\ &= 2\pi \left( {\frac{{27}}{4}} \right)\\ &= \frac{{27}}{2}\pi \end{aligned}\)

Therefore, the volume of the solid is \(\frac{{27}}{2}\pi .\)