009B Sample Final 2, Problem 5

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

(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 surface area  $S$ of a function  $y=f(x)$ rotated about the  $y$ -axis is given by

$S=\int 2\pi x\,ds,$ where  $ds={\sqrt {1+{\bigg (}{\frac {dx}{dy}}{\bigg )}^{2}}}dy.$ 2. The formula for the length  $L$ of a curve  $y=f(x)$ where  $a\leq x\leq b$ is

$L=\int _{a}^{b}{\sqrt {1+{\bigg (}{\frac {dy}{dx}}{\bigg )}^{2}}}~dx.$ 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
${\begin{array}{rcl}\displaystyle {S}&=&\displaystyle {\int _{0}^{1}2\pi x{\sqrt {1+(3y^{2})^{2}}}~dy}\\&&\\&=&\displaystyle {2\pi \int _{0}^{1}y^{3}{\sqrt {1+9y^{4}}}~dy.}\end{array}}$ where  $S$ is the surface area.
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
$L=\int _{1}^{4}{\sqrt {1+{\frac {729x}{4}}}}~dx.$ Now, we use  $u$ -substitution.
Let  $u=1+{\frac {729x}{4}}.$ Then,  $du={\frac {729}{4}}~dx$ and  $dx={\frac {4}{729}}~du.$ Also, since this is a definite integral, we need to change the bounds of integration.
We have
$u_{1}=1+{\frac {729(1)}{4}}={\frac {733}{4}}$ and  $u_{2}=1+{\frac {729(4)}{4}}=730.$ Hence, we now have
$L=\int _{\frac {733}{4}}^{730}{\frac {4}{729}}u^{\frac {1}{2}}~du.$ Step 3:
Therefore, we have
${\begin{array}{rcl}\displaystyle {L}&=&\displaystyle {{\frac {4}{729}}{\bigg (}{\frac {2}{3}}u^{\frac {3}{2}}{\bigg )}{\bigg |}_{\frac {733}{4}}^{730}}\\&&\\&=&\displaystyle {{\frac {8}{2187}}u^{\frac {3}{2}}{\bigg |}_{\frac {733}{4}}^{730}}\\&&\\&=&\displaystyle {{\frac {8}{2187}}(730)^{\frac {3}{2}}-{\frac {8}{2187}}{\bigg (}{\frac {733}{4}}{\bigg )}^{\frac {3}{2}}.}\end{array}}$ (a)    ${\frac {\pi }{27}}(10)^{\frac {3}{2}}-{\frac {\pi }{27}}$ (b)    ${\frac {8}{2187}}(730)^{\frac {3}{2}}-{\frac {8}{2187}}{\bigg (}{\frac {733}{4}}{\bigg )}^{\frac {3}{2}}$ 