Answer to Question #114610 in Calculus for ANJU JAYACHANDRAN

Question

Find the area of the surface that is generated by revolving the portion of the curve y=x^2 between x=1 and x=2 about the y-axis.

Answer 1

The curve y = $\ x^{2}$ from x = 1

x=1 and x = 2

x=2 about the y axis

We know that x = 1 to x = 2 , which is analogous from y = 1 to y = 4.

Area of the surface is

$A = 2\times \pi \int_{a}^{b} g(y)\sqrt{1+(g'(y)^{2})}dy$

$g(y) = \sqrt{y}$

$g'(y) = \frac {1}{(2\times\sqrt{y})}$

$A = 2\times \pi \int_{1}^{4} \sqrt {y} \sqrt{1+(\frac{1}{(2\times\sqrt{y})})^{2}}dy$

$A = 2\times \pi \int_{1}^{4} \sqrt{y(1+\frac{1}{4y})}dy$

$A = 2\times \pi \int_{1}^{4} \sqrt{y+(\frac{1}{4})}dy$

Let,

$u = y + \frac {1}{4}$

$du = dy$

$y = 1 , u = \frac {5} {4}$

$y = 1 , u = \frac {17} {4} ​$

$A = 2\times \pi \int_{\frac{5}{4}}^{\frac{17}{4}} u^ { \frac {1} {2} } dy$

$A= 2 \times \pi [\frac{2}{3} (u^\frac {3} {2}) ]|_{\frac{5}{4}} ^{ \frac{17}{4}}$

$A= \frac{4 \times \pi}{3} [(\frac{17}{4})^\frac {3} {2} - (\frac{5}{4})^\frac {3} {2}]$

$A= 30.85 sq.unit$