Question

Find the area of the surface that is generated by revolving the portion of the curve y=x^2 between x=0 and x=1 about the y-axis.

Answer 1

Solution:

The curve y = $\ x^{2}$ from x = 0 and x = 1 about the y axis

We know that x = 0 to x = 1 , which is analogous from y = 0 to y = 1.

Area of the surface is

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

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

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

$A = 2\times \pi \int_{0}^{1} \sqrt {y} \sqrt{1+(\frac{1}{(2\times\sqrt{y})})^{2}}dy$

$\Rightarrow A = 2\times \pi \int_{0}^{1} \sqrt{y(1+\frac{1}{4y})}dy$

$\Rightarrow A = 2\times \pi \int_{0}^{1} \sqrt{y+(\frac{1}{4})}dy$

Let,

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

$\Rightarrow du = dy$

$y = 0 , u = \frac {1} {4}$

$y = 1 , u = \frac {5} {4} ​$

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

$\Rightarrow A= 2 \times \pi [\frac{2}{3} (u^\frac {3} {2}) ]|_{\frac{1}{4}} ^{ \frac{5}{4}}$

$\Rightarrow A= \frac{4 \times \pi}{3} [(\frac{5}{4})^\frac {3} {2} - (\frac{1}{4})^\frac {3} {2}]$

$\Rightarrow A= 5.33 sq.unit$