Definite Integration Question 45
Question: The part of straight line $ y=x+1 $ between $ x=2 $ and $ x=3 $ is revolved about x-axis, then the curved surface of the solid thus generated is
[UPSEAT 2000]
Options:
A) $ 37\pi /3 $
B) $ 7\pi \sqrt{2} $
C) $ 37\pi $
D) $ y=x^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
Curved surface $ =\int_a^{b}{2\pi y\sqrt{[ 1+{{( \frac{dy}{dx} )}^{2}} ]}dx} $
Given that $ a=2, $
$ b=3 $ and $ y=x+1 $ .
On differentiating with respect to $ x $ ,
$ \frac{dy}{dx}=1+0or\frac{dy}{dx}=1 $
Therefore, curved surface $ =\int_2^{3}{2\pi (x+1)\sqrt{[1+{{(1)}^{2}}]}dx} $
$ =\int_2^{3}{2\pi (x+1)\sqrt{2}}dx $
$ =2\sqrt{2}\pi \int_2^{3}{(x+1)}dx $
$ =2\sqrt{2}\pi [ \frac{{{(x+1)}^{2}}}{2} ]_2^{3} $
$ =\frac{2\sqrt{2}}{2}\pi [{{(3+1)}^{2}}-{{(2+1)}^{2}}]=\sqrt{2}\pi (16-9)=7\sqrt{2}\pi =7\pi \sqrt{2} $ .
BETA
