Answer:
[tex]\mathbf{ \dfrac{\pi}{675}\Big[ 34\sqrt{34} -125\Big] }[/tex]
Step-by-step explanation:
The curve x = f(y)
The area of the surface around the y-axis from y = a → y = b is:
[tex]=\int^b_a 2 \pi x \sqrt{1 + (\dfrac{dx}{dy})^2} \ dy[/tex]
From the given curve:
[tex]y = (5x)^{^{\dfrac{1}{3}}[/tex] ; assuming the region bounded by the curve is 0 ≤ y ≤ 1
So;
[tex]y = (5x)^{^{\dfrac{1}{3}}[/tex]
5x = y³
[tex]x = \dfrac{1}{5}y^3[/tex]
The differential of the above equation Is:
[tex]\dfrac{dx}{dy}= \dfrac{1}{5} \times (3y^2)[/tex]
[tex]\dfrac{dx}{dy}= \dfrac{3}{5}y^2[/tex]
Now, we have the area of the surface produced around the curve [tex]x = \dfrac{1}{5}y^3[/tex] through the y axis from the region y = 0 to y = 1
∴
[tex]= \int ^1_0 2 \pi \dfrac{1}{5}y^3 \sqrt{1 + (\dfrac{3}{5}y^2)^2} \ dy[/tex]
[tex]= \dfrac{ 2 \pi}{5} \int ^1_0 y^3 \sqrt{1 + \dfrac{9}{25}y^4} \ dy[/tex]
[tex]= \dfrac{ 2 \pi}{5} \int ^1_0 y^3 \sqrt{ \dfrac{25+9y^4}{25}} \ dy[/tex]
[tex]= \dfrac{ 2 \pi}{5} \int ^1_0 y^3 \dfrac{\sqrt{25+9y^4}}{5}} \ dy[/tex]
[tex]= \dfrac{ 2 \pi}{25} \int ^1_0 y^3 \sqrt{25+9y^4}} \ dy[/tex]
Let make [tex]u = \sqrt{25+9y^4}[/tex]
It implies that:
[tex]u^3 = (25+9y^4)\sqrt{25+9y^4}[/tex]
[tex]u = \sqrt{25+9y^4} \\\\ du = \dfrac{1}{2\sqrt{25 +9y^4}}(36y^3) \ dy[/tex]
[tex]du = \dfrac{18y^3}{\sqrt{25 +9y^4}}\ dy[/tex]
[tex]y^3dy = \dfrac{1}{18}\sqrt{25+9y^4} \ du[/tex]
when y = 0 ;
[tex]u = \sqrt{25+ 9(0)^4}[/tex]
[tex]u = \sqrt{25}[/tex]
u = 5
when y = 1;
[tex]u = \sqrt{25+ 9(1)^4}[/tex]
[tex]u = \sqrt{25+9}[/tex]
[tex]u = \sqrt{34}[/tex]
∴
The equation [tex]\dfrac{ 2 \pi}{25} \int ^1_0 y^3 \sqrt{25+9y^4}} \ dy[/tex] can be written as:
[tex]= \dfrac{2 \pi}{25} \int ^{\sqrt{34}}_{5} (u ) \dfrac{1}{18} \ udu[/tex]
[tex]= \dfrac{2 \pi}{25\times 18} \int ^{\sqrt{34}}_{5} (u ) \ udu[/tex]
[tex]= \dfrac{\pi}{225} \int ^{\sqrt{34}}_{5} (u^2 ) \ udu[/tex]
[tex]= \dfrac{\pi}{225}\Big[ \dfrac{u^3}{3} \Big] ^{\sqrt{34}}_{3}\\[/tex]
[tex]\mathbf{= \dfrac{\pi}{675}\Big[ 34\sqrt{34} -125\Big] }[/tex]
