Question

Find the area of the surface generated when the given curve is revolved about the given axis. 5x^1/3

Answer 1

Answer:

$\mathbf{ \dfrac{\pi}{675}\Big[ 34\sqrt{34} -125\Big] }$

Step-by-step explanation:

The curve x = f(y)

The area of the surface around the y-axis from y = a → y = b is:

$=\int^b_a 2 \pi x \sqrt{1 + (\dfrac{dx}{dy})^2} \ dy$

From the given curve:

$y = (5x)^{^{\dfrac{1}{3}}$ ; assuming the region bounded by the curve is 0 ≤ y ≤ 1

So;

$y = (5x)^{^{\dfrac{1}{3}}$

5x = y³

$x = \dfrac{1}{5}y^3$

The differential of the above equation Is:

$\dfrac{dx}{dy}= \dfrac{1}{5} \times (3y^2)$

$\dfrac{dx}{dy}= \dfrac{3}{5}y^2$

Now, we have the area of the surface produced around the curve $x = \dfrac{1}{5}y^3$ through the y axis from the region y = 0 to y = 1

∴

$= \int ^1_0 2 \pi \dfrac{1}{5}y^3 \sqrt{1 + (\dfrac{3}{5}y^2)^2} \ dy$

$= \dfrac{ 2 \pi}{5} \int ^1_0 y^3 \sqrt{1 + \dfrac{9}{25}y^4} \ dy$

$= \dfrac{ 2 \pi}{5} \int ^1_0 y^3 \sqrt{ \dfrac{25+9y^4}{25}} \ dy$

$= \dfrac{ 2 \pi}{5} \int ^1_0 y^3 \dfrac{\sqrt{25+9y^4}}{5}} \ dy$

$= \dfrac{ 2 \pi}{25} \int ^1_0 y^3 \sqrt{25+9y^4}} \ dy$

Let make $u = \sqrt{25+9y^4}$

It implies that:

$u^3 = (25+9y^4)\sqrt{25+9y^4}$

$u = \sqrt{25+9y^4} \\\\ du = \dfrac{1}{2\sqrt{25 +9y^4}}(36y^3) \ dy$

$du = \dfrac{18y^3}{\sqrt{25 +9y^4}}\ dy$

$y^3dy = \dfrac{1}{18}\sqrt{25+9y^4} \ du$

when y = 0 ;

$u = \sqrt{25+ 9(0)^4}$

$u = \sqrt{25}$

u = 5

when y = 1;

$u = \sqrt{25+ 9(1)^4}$

$u = \sqrt{25+9}$

$u = \sqrt{34}$

∴

The equation $\dfrac{ 2 \pi}{25} \int ^1_0 y^3 \sqrt{25+9y^4}} \ dy$ can be written as:

$= \dfrac{2 \pi}{25} \int ^{\sqrt{34}}_{5} (u ) \dfrac{1}{18} \ udu$

$= \dfrac{2 \pi}{25\times 18} \int ^{\sqrt{34}}_{5} (u ) \ udu$

$= \dfrac{\pi}{225} \int ^{\sqrt{34}}_{5} (u^2 ) \ udu$

$= \dfrac{\pi}{225}\Big[ \dfrac{u^3}{3} \Big] ^{\sqrt{34}}_{3}$

$\mathbf{= \dfrac{\pi}{675}\Big[ 34\sqrt{34} -125\Big] }$