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 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
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{\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  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 \frac{dy}{dx}.}
Since  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 y=1+9x^{\frac{3}{2}},}   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 \frac{dy}{dx}=\frac{27\sqrt{x}}{2}.}
Then, the arc length  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 L}   of the curve is given by
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 L=\int_1^4 \sqrt{1+\bigg(\frac{27\sqrt{x}}{2}\bigg)^2}~dx.}

Step 2:
Then, 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 L=\int_1^4 \sqrt{1+\frac{27^2x}{2^2}}~dx.}
Now, we use  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 u} -substitution.
Let  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 u=1+\frac{27^2x}{2^2}.}
Then,  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 du=\frac{27^2}{2^2}dx}   and  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 dx=\frac{2^2}{27^2}du.}
Also, since this is a definite integral, we need to change the bounds of integration.
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 u_1=1+\frac{27^2(1)}{2^2}=1+\frac{27^2}{2^2}} and  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 u_2=1+\frac{27^2(4)}{2^2}=1+27^2.}
Hence, we now 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 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
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{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)    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 \frac{\pi}{27} (10)^{\frac{3}{2}}-\frac{\pi}{27}}
(b)    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 \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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