Question

the area of the surface bounded by rotating the arc of a parabola y=x^2 from (1,1),(2,4) about y axis

Answer 1

Answer:$\frac{1}{6} \pi ( 17^{\frac{3}{2}} - 5^{\frac{3}{2}} )$

Step-by-step explanation:

We have $x = \sqrt{y}$

By differentiating with respect to y,

$\frac{dx}{dy} = \frac{1}{2} y^{-\frac{1}{2}$

By squaring,

$(\frac{dx}{dy})^2 = \frac{1}{4} y^{-1}$

Thus, the area of the surface bounded by rotating the arc of a parabola $y=x^2$ from (1,1),(2,4) about y axis,

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

$= \pi [ \frac{2}{12} (4y+1)^{\frac{3}{2}} ]_{1}^{4}$

$= \frac{1}{6} \pi ( 17^{\frac{3}{2}} - 5^{\frac{3}{2}} )$