009B Sample Final 2, Problem 5

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

$y^{3}=x$

between $(0,0)$ and $(1,1)$ about the $y$-axis.

(b) Find the length of the arc

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

between the points $(1,10)$ and $(4,73).$

Foundations:
1. The formula for the length  $L$  of a curve  $y=f(x)$  where  $a\leq x\leq b$  is
$L=\int _{a}^{b}{\sqrt {1+{\bigg (}{\frac {dy}{dx}}{\bigg )}^{2}}}~dx.$
2. The surface area  $S$  of a function  $y=f(x)$  rotated about the  $y$-axis is given by
$S=\int 2\pi x\,ds,$  where $ds={\sqrt {1+{\bigg (}{\frac {dx}{dy}}{\bigg )}^{2}}}dy.$

Solution:

(a)

Step 1:
We start by calculating  ${\frac {dx}{dy}}.$
Since  $x=y^{3},~{\frac {dx}{dy}}=3y^{2}.$
Now, we are going to integrate with respect to $y.$
Using the formula given in the Foundations section,
we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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.} |- |where &nbsp;<math>S}   is the surface area. \end{array}</math>

Step 2:
Now, we use $u$-substitution.
Let $u=1+9y^{4}.$
Then, $du=36y^{3}dy$ and ${\frac {du}{36}}=y^{3}dy.$
Also, since this is a definite integral, we need to change the bounds of integration.
We have
$u_{1}=1+9(0)^{4}=1$ and $u_{2}=1+9(1)^{4}=10.$
Thus, we get
${\begin{array}{rcl}\displaystyle {S}&=&\displaystyle {{\frac {2\pi }{36}}\int _{1}^{10}{\sqrt {u}}~du}\\&&\\&=&\displaystyle {{\frac {\pi }{27}}u^{\frac {3}{2}}{\bigg |}_{1}^{10}}\\&&\\&=&\displaystyle {{\frac {\pi }{27}}(10)^{\frac {3}{2}}-{\frac {\pi }{27}}.}\end{array}}$

(b)

Step 1:
First, we calculate  ${\frac {dy}{dx}}.$
Since $y=1+9x^{\frac {3}{2}},$ we have
${\frac {dy}{dx}}={\frac {27{\sqrt {x}}}{2}}.$
Then, the arc length  $L$  of the curve is given by
$L=\int _{1}^{4}{\sqrt {1+{\bigg (}{\frac {27{\sqrt {x}}}{2}}{\bigg )}^{2}}}~dx.$

Step 2:
Then, we have
$L=\int _{1}^{4}{\sqrt {1+{\frac {27^{2}x}{2^{2}}}}}~dx.$
Now, we use  $u$-substitution.
Let  $u=1+{\frac {27^{2}x}{2^{2}}}.$
Then,   $du={\frac {27^{2}}{2^{2}}}dx$  and  $dx={\frac {2^{2}}{27^{2}}}du.$
Also, since this is a definite integral, we need to change the bounds of integration.
We have   $u_{1}=1+{\frac {27^{2}(1)}{2^{2}}}=1+{\frac {27^{2}}{2^{2}}}$
and   $u_{2}=1+{\frac {27^{2}(4)}{2^{2}}}=1+27^{2}.$
Hence, we now have
$L=\int _{1+{\frac {27^{2}}{2^{2}}}}^{1+27^{2}}{\frac {2^{2}}{27^{2}}}u^{\frac {1}{2}}~du.$

Step 3:
Therefore, we have
${\begin{array}{rcl}\displaystyle {L}&=&\displaystyle {{\frac {2^{2}}{27^{2}}}{\bigg (}{\frac {2}{3}}u^{\frac {3}{2}}{\bigg )}{\bigg |}_{1+{\frac {27^{2}}{2^{2}}}}^{1+27^{2}}}\\&&\\&=&\displaystyle {{\frac {2^{3}}{3^{4}}}u^{\frac {3}{2}}{\bigg |}_{1+{\frac {27^{2}}{2^{2}}}}^{1+27^{2}}}\\&&\\&=&\displaystyle {{\frac {2^{3}}{3^{4}}}(1+27^{2})^{\frac {3}{2}}-{\frac {2^{3}}{3^{4}}}{\bigg (}1+{\frac {27^{2}}{2^{2}}}{\bigg )}^{\frac {3}{2}}.}\end{array}}$

Final Answer:
(a)    ${\frac {\pi }{27}}(10)^{\frac {3}{2}}-{\frac {\pi }{27}}$
(b)    ${\frac {2^{3}}{3^{4}}}(1+27^{2})^{\frac {3}{2}}-{\frac {2^{3}}{3^{4}}}{\bigg (}1+{\frac {27^{2}}{2^{2}}}{\bigg )}^{\frac {3}{2}}$

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