# 009B Sample Final 2, Problem 5

(a) Find the area of the surface obtained by rotating the arc of the curve

${\displaystyle y^{3}=x}$

between  ${\displaystyle (0,0)}$  and  ${\displaystyle (1,1)}$  about the  ${\displaystyle y}$-axis.

(b) Find the length of the arc

${\displaystyle y=1+9x^{\frac {3}{2}}}$

between the points  ${\displaystyle (1,10)}$  and  ${\displaystyle (4,73).}$

Foundations:
1. The surface area  ${\displaystyle S}$  of a function  ${\displaystyle y=f(x)}$  rotated about the  ${\displaystyle y}$-axis is given by

${\displaystyle S=\int 2\pi x\,ds,}$  where  ${\displaystyle ds={\sqrt {1+{\bigg (}{\frac {dx}{dy}}{\bigg )}^{2}}}dy.}$

2. The formula for the length  ${\displaystyle L}$  of a curve  ${\displaystyle y=f(x)}$  where  ${\displaystyle a\leq x\leq b}$  is

${\displaystyle L=\int _{a}^{b}{\sqrt {1+{\bigg (}{\frac {dy}{dx}}{\bigg )}^{2}}}~dx.}$

Solution:

(a)

Step 1:
We start by calculating  ${\displaystyle {\frac {dx}{dy}}.}$
Since  ${\displaystyle x=y^{3},}$
${\displaystyle {\frac {dx}{dy}}=3y^{2}.}$
Now, we are going to integrate with respect to  ${\displaystyle y.}$
Using the formula given in the Foundations section,
we have
${\displaystyle {\begin{array}{rcl}\displaystyle {S}&=&\displaystyle {\int _{0}^{1}2\pi x{\sqrt {1+(3y^{2})^{2}}}~dy}\\&&\\&=&\displaystyle {2\pi \int _{0}^{1}y^{3}{\sqrt {1+9y^{4}}}~dy.}\end{array}}}$
where  ${\displaystyle S}$  is the surface area.
Step 2:
Now, we use  ${\displaystyle u}$-substitution.
Let  ${\displaystyle u=1+9y^{4}.}$
Then,  ${\displaystyle du=36y^{3}dy}$  and  ${\displaystyle {\frac {du}{36}}=y^{3}dy.}$
Also, since this is a definite integral, we need to change the bounds of integration.
We have
${\displaystyle u_{1}=1+9(0)^{4}=1}$  and  ${\displaystyle u_{2}=1+9(1)^{4}=10.}$
Thus, we get
${\displaystyle {\begin{array}{rcl}\displaystyle {S}&=&\displaystyle {{\frac {2\pi }{36}}\int _{1}^{10}{\sqrt {u}}~du}\\&&\\&=&\displaystyle {{\frac {\pi }{27}}u^{\frac {3}{2}}{\bigg |}_{1}^{10}}\\&&\\&=&\displaystyle {{\frac {\pi }{27}}(10)^{\frac {3}{2}}-{\frac {\pi }{27}}.}\end{array}}}$

(b)

Step 1:
First, we calculate  ${\displaystyle {\frac {dy}{dx}}.}$
Since  ${\displaystyle y=1+9x^{\frac {3}{2}},}$  we have
${\displaystyle {\frac {dy}{dx}}={\frac {27{\sqrt {x}}}{2}}.}$
Then, the arc length  ${\displaystyle L}$  of the curve is given by
${\displaystyle L=\int _{1}^{4}{\sqrt {1+{\bigg (}{\frac {27{\sqrt {x}}}{2}}{\bigg )}^{2}}}~dx.}$
Step 2:
Then, we have
${\displaystyle L=\int _{1}^{4}{\sqrt {1+{\frac {729x}{4}}}}~dx.}$
Now, we use  ${\displaystyle u}$-substitution.
Let  ${\displaystyle u=1+{\frac {729x}{4}}.}$
Then,  ${\displaystyle du={\frac {729}{4}}~dx}$  and  ${\displaystyle dx={\frac {4}{729}}~du.}$
Also, since this is a definite integral, we need to change the bounds of integration.
We have
${\displaystyle u_{1}=1+{\frac {729(1)}{4}}={\frac {733}{4}}}$  and  ${\displaystyle u_{2}=1+{\frac {729(4)}{4}}=730.}$
Hence, we now have
${\displaystyle L=\int _{\frac {733}{4}}^{730}{\frac {4}{729}}u^{\frac {1}{2}}~du.}$
Step 3:
Therefore, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {L}&=&\displaystyle {{\frac {4}{729}}{\bigg (}{\frac {2}{3}}u^{\frac {3}{2}}{\bigg )}{\bigg |}_{\frac {733}{4}}^{730}}\\&&\\&=&\displaystyle {{\frac {8}{2187}}u^{\frac {3}{2}}{\bigg |}_{\frac {733}{4}}^{730}}\\&&\\&=&\displaystyle {{\frac {8}{2187}}(730)^{\frac {3}{2}}-{\frac {8}{2187}}{\bigg (}{\frac {733}{4}}{\bigg )}^{\frac {3}{2}}.}\end{array}}}$

(a)    ${\displaystyle {\frac {\pi }{27}}(10)^{\frac {3}{2}}-{\frac {\pi }{27}}}$
(b)    ${\displaystyle {\frac {8}{2187}}(730)^{\frac {3}{2}}-{\frac {8}{2187}}{\bigg (}{\frac {733}{4}}{\bigg )}^{\frac {3}{2}}}$