Find the area of the surface generaled by revolving the curve x=y35, 0≤y≤3 about the y-axis.

Step 1The curve x=g(y), c≤y≤d is revolved about the x-axis, then area of the surface isA=2π∫cdg(y)1+[g′(y)]2dyStep 2Now differentiate g(y)=y32, with resprct to y, we getg′(y)=3y25Step 3Now substitute the value of g(y) and g′(y) in above formula, we getA=2π∫03y351+(3y25)2dyStep 4Simplify the above integral, we getA=2π∫03y351+(3y25)2dyA=2π5∫03y31+9y425dyPut 1+9y425=u⇒y3dy=2536dyWhen y=0, u=1 and when y=3, u=75425So, A=2π52536∫174525uduA=5π18[u3232]175425A=5π18[(75425)32−1]A=5π27[165.63−1]A=30.49π

Solved: "Find the area of the surface generaled by..."

nemired9

Answered question

2021-12-11

Find the area of the surface generaled by revolving the curve

x = \frac{y^{3}}{5}, 0 \leq y \leq 3

about the y-axis.

Answer & Explanation

Elois Puryear

Beginner2021-12-12Added 30 answers

Step 1
The curve $x = g (y), c \leq y \leq d$ is revolved about the x-axis, then area of the surface is
$A = 2 π \int_{c}^{d} g (y) \sqrt{1 + {[g^{'} (y)]}^{2}} d y$
Step 2
Now differentiate $g (y) = y^{\frac{3}{2}}$ , with resprct to y, we get
$g^{'} (y) = \frac{3 y^{2}}{5}$
Step 3
Now substitute the value of g(y) and $g^{'} (y)$ in above formula, we get
$A = 2 π \int_{0}^{3} \frac{y^{3}}{5} \sqrt{1 + {(\frac{3 y^{2}}{5})}^{2}} d y$
Step 4
Simplify the above integral, we get
$A = 2 π \int_{0}^{3} \frac{y^{3}}{5} \sqrt{1 + {(\frac{3 y^{2}}{5})}^{2}} d y$
$A = \frac{2 π}{5} \int_{0}^{3} y^{3} \sqrt{1 + \frac{9 y^{4}}{25}} d y$
Put $1 + \frac{9 y^{4}}{25} = u \Rightarrow y^{3} d y = \frac{25}{36} d y$
When $y = 0, u = 1$ and when $y = 3, u = \frac{754}{25}$
So, $A = \frac{2 π}{5} \frac{25}{36} \int_{1}^{\frac{745}{25}} \sqrt{u} d u$
$A = \frac{5 π}{18} {[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}]}_{1}^{\frac{754}{25}}$
$A = \frac{5 π}{18} [{(\frac{754}{25})}^{\frac{3}{2}} - 1]$
$A = \frac{5 π}{27} [165.63 - 1]$
$A = 30.49 π$

enlacamig

Beginner2021-12-13Added 30 answers

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