Q 57 Use definite integrals to find t... [FREE SOLUTION]

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Chapter 6: Q 57 (page 524)

Use definite integrals to find the volume of each solid of revolution described in Exercises $49 - 61$ . (It is your choice whether to use disks/washers or shells in these exercises.)
The region between the graph of $f (x) = x^{2} - 4 x + 4$ and the $x -$ axis on role="math" localid="1651328870762" $[0, 2]$ , revolved around the $y -$ axis.

Short Answer

Expert verified

The required volume by using shells is $V = \frac{8 蟺}{3}$ .

Step by step solution

Step 1. Given Information

We have given a function :-

$f (x) = x^{2} - 4 x + 4$

We have to find the volume of region of graph of this function and $x - a x i s$ on $[0, 2]$ revolved around $y - a x i s$ .

Find the integral and evaluate it to calculate volume

We know that by using shells the volume is given by :-

$V = 2 蟺 {鈭�}_{c}^{d} r (x) h (x) dx$

Here axis of revolution is $y - a x i s$ . So radius is $r (x) = x$ and height is given by the function.

So $h (x) = x^{2} - 4 x + 4$ .

Also the limits will be $0 t o 2$ .

Then we get the volume as following :-

$V = 2 蟺 {鈭�}_{0}^{2} x (x^{2} - 4 x + 4) dx 鈬� V = 2 蟺 {鈭�}_{0}^{2} (x^{3} - 4 x^{2} + 4 x) dx 鈬� V = 2 蟺 {[\frac{x^{4}}{4} - \frac{4 x^{3}}{3} + \frac{4 x^{2}}{2}]}_{0}^{2} 鈬� V = 2 蟺 (\frac{16}{4} - \frac{32}{3} + \frac{16}{2}) - 0 鈬� V = 2 蟺 (4 - \frac{32}{3} + 8) 鈬� V = 2 蟺 (12 - \frac{32}{3}) 鈬� V = 2 蟺 (\frac{36 - 32}{3}) 鈬� V = 2 蟺脳 \frac{4}{3} 鈬� V = \frac{8 蟺}{3}$