Question

Find the exact area of the surface obtained by rotating the curve about the x-axis. y = sin πx 3 , 0 ≤ x ≤ 3

Answer 1

Answer:

$\mathbf{S =6 \sqrt{1 + \dfrac{\pi^2}{9} }+ \dfrac{18}{\pi} In (\dfrac{\pi}{3}+ \sqrt{1+ \dfrac{\pi^2}{9}})}$

Step-by-step explanation:

Given that

curve $y = \dfrac{\pi x}{3}, 0 \leq x \leq 3$

The objective is to find the area of the surface obtained by rotating the above curve about the x-axis.

Suppose f is positive and posses a continuous derivative,

the surface is gotten by the rotating the curve about the x-axis is:

$S = \int ^b_a 2 \pi f (x) \sqrt {1 + (f' (x))^2 } \ dx$

The derivative of the function $y' = \dfrac{\pi}{3} cos \dfrac{\pi x}{3}$

As such, the surface area is:

$S = \int ^3_0 2 \pi sin \dfrac{\pi x}{3} \sqrt {1 +(\dfrac{\pi}{3}cos \dfrac{\pi x}{3})^2 } \ dx$

Suppose ;

$u = \dfrac{\pi}{3}cos \dfrac{\pi x}{3}$

$du = -( \dfrac{\pi}{3})^2 sin \dfrac{\pi x}{3} \ dx$

If x = 0 , then $u = \dfrac{\pi}{3}cos \dfrac{\pi (0)}{3} = \dfrac{\pi}{3}$

If x = 3 , then $u = \dfrac{\pi}{3}cos \dfrac{\pi (3)}{3}$

$u = \dfrac{\pi}{3}(-1)$

$u = -\dfrac{\pi}{3}$

The equation for S can now be rewritten as:

$S = \int^3_0 2 \pi sin \dfrac{\pi x}{3} \sqrt{1+(\dfrac{\pi}{3} cos \dfrac{\pi x}{3})^2 }\ dx$

$S = 2 \pi \int ^{-\frac{\pi}{3} }_{\frac{\pi}{3}}(-\dfrac{9 \ du }{\pi^2} ) \sqrt{1+u^2}$

$S = 18 \pi * \dfrac{1}{\pi ^2 } \int ^{-\frac{\pi}{3}}_{\frac{\pi}{3}} \sqrt{1+u^2} \ du$

$S = \dfrac{18} {\pi} \int ^{-\frac{\pi}{3}}_{\frac{\pi}{3}} \sqrt{1+u^2} \ du$

$S = \dfrac{18} {\pi} (2 \int ^{-\frac{\pi}{3}}_{0} \sqrt{1+u^2} \ du)$

since $(\int ^a_{-a} fdx = 2\int^a_0 fdx , f= \sqrt{1+u^2} \ is \ even )$

Applying the formula:

$\int {\sqrt{1+x^2}} \ d x= \dfrac{x}{2} \sqrt{1+x^2}+ \dfrac{1}{2} In ( x + \sqrt{1+x^2})$

$S = \dfrac{36}{x}[ \dfrac{u}{2} \sqrt{1+u^2}+ \dfrac{1}{2} \ In (u+ \sqrt{1+u^2}) ] ^{\frac{\pi}{3}}_{0}$

$S = \dfrac{36}{x}[ \dfrac{\dfrac{\pi}{3}}{2} \sqrt{1+\dfrac{\pi^2}{9}}+ \dfrac{1}{2} \ In (\dfrac{\pi}{3}+ \sqrt{1+\dfrac{\pi^2}{9}})-0 ]$

$S =6 \sqrt{1 + \dfrac{\pi^2}{9} }+ \dfrac{18}{\pi} In (\dfrac{\pi}{3}+ \sqrt{1+ \dfrac{\pi^2}{9}})$

Therefore, the exact area of the surface is $\mathbf{S =6 \sqrt{1 + \dfrac{\pi^2}{9} }+ \dfrac{18}{\pi} In (\dfrac{\pi}{3}+ \sqrt{1+ \dfrac{\pi^2}{9}})}$

Answer 2

The area of the surface is,$S = 6\sqrt{1+\dfrac{\pi ^2}{3}} + \dfrac{18}{3}\ ln (\dfrac{\pi }{3}+\sqrt{1+\dfrac{\pi ^2} {9}})$.

Given that,

The exact area of the surface obtained by rotating the curve about the x-axis. y = sin πx\3 , 0 ≤ x ≤ 3.

We have to determine,

Area of the surface obtained by rotating the curve.

According to the question,

Suppose f is positive and posses a continuous derivative,

The surface is gotten by the rotating the curve about the x-axis is:

Area of the surface is given by,

$S = \int\limits^b_a {2\pi f(x) .\sqrt{1+ (f'(x))} } \, dx$

The given curve is x-axis,

$y = \dfrac{sin\pi x}{3}$

The derivative of the function is,

$\dfrac{dy}{dx} =\dfrac{\pi }{3} \dfrac{ cos\pi x}{3}$

The surface area is,

$S = \int\limits^b_a {2\pi f(x) .\sqrt{1+(\dfrac{\pi }{3} \dfrac{ cos\pi x}{3})^2} }}} \, dx$

Substitute the value of f(x),

$S = \int\limits^3_0{2\pi\dfrac{sin\pi x}{3} .\sqrt{1+(\dfrac{\pi }{3} \dfrac{ cos\pi x}{3})^2} }}} \, dx$

Suppose;

$u = \dfrac{\pi }{3}cos\dfrac{\pi x}{3}dx\\\\\du =( \dfrac{-\pi }{3})^2 sin\dfrac{\pi x}{3}dx\\\\if \ x = 0, \ then \ u = \dfrac{\pi }{3}cos\dfrac{\pi (0)}{3} = \dfrac{\pi }{3}\\\\if \ x = 3, \ then \ u = \dfrac{\pi }{3}cos\dfrac{\pi (3)}{3} = \dfrac{\pi }{3}(-1)= \dfrac{-\pi }{3}$

Then,

$S = \int\limits^3_0{2\pi\dfrac{sin\pi x}{3} .\sqrt{1+(\dfrac{\pi }{3} \dfrac{ cos\pi x}{3})^2} }}} \, dx\\\\S = 2\pi \int\limits^{\frac{-\pi}{3}}_ \frac{\pi }{3} (\dfrac{-9du}{\pi ^2})\sqrt{1+u^2} \ du\\\\S = 18\pi \times \dfrac{1}{\pi ^2}\int\limits^{\frac{-\pi}{3}}_ \frac{\pi }{3} \sqrt{1+u^2} \ du\\\\S = \dfrac{18}{\pi}\int\limits^{\frac{-\pi}{3}}_ {0} 2\sqrt{1+u^2} \ du$

Since,

$(\int\limits^a_{-a}{f} \, dx = 2\int\limits^a_0 {f} \, dx , \ f = \sqrt{1+u^2}\ is \ even)$

By applying the formula to solve the given integration,

$\int {\sqrt{1+x^2} } \, dx = \dfrac{x}{2}\sqrt{1+x^2} + \dfrac{1}{2} \ ln(x+\sqrt{1+x^2})\\\\S = \dfrac{36}{2} [ \dfrac{u}{2} \sqrt{1+u^2} + \dfrac{1}{2} \ ln(u+\sqrt{1+u^2})}]^{\frac{\pi }{3}}_0$

$S = \dfrac{36}{2} [ \dfrac{\dfrac{\pi }{3}}{2} \sqrt{1+\dfrac{\pi^2 }{9}} + \dfrac{1}{2} \ ln(\dfrac{\pi }{3}+\sqrt{1+\dfrac{\pi ^2} {9}}-0)]\\\\S = 6\sqrt{1+\dfrac{\pi ^2}{3}} + \dfrac{18}{3}\ ln (\dfrac{\pi }{3}+\sqrt{1+\dfrac{\pi ^2} {9}})$

Hence, The area of the surface is,$S = 6\sqrt{1+\dfrac{\pi ^2}{3}} + \dfrac{18}{3}\ ln (\dfrac{\pi }{3}+\sqrt{1+\dfrac{\pi ^2} {9}})$

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