Surface Area Solution 9

Processing math: 100%

Solution 9: First solve the equation

$x^{2/3} + y^{2/3}=1$ for

$x$ getting

$x^{2/3} = 1 - y^{2/3} \ \ \ \longrightarrow$

$x = (1 - y^{2/3})^{3/2}$ Here is a carefully labeled sketch of the graph with a radius

$r$ marked together with

$y$ on the

$y$ -axis. We will find the surface area of the TOP HALF of theSurface of Revolution and DOUBLE our answer.

tex2html_wrap_inline125

The derivative of

$x$ is

${ dx \over dy } = {3 \over 2} (1-y^{2/3})^{1/2} \cdot {-2 \over 3}y^{-1/3} \ \ \ \longrightarrow$

${ dx \over dy } = { -(1-y^{2/3})^{1/2} \over y^{1/3} }$
Thus the total Area of this Surface of Revolution is

$Surface \ Area = 2 \pi \cdot 2 \int_{0}^{1} (radius) \sqrt{ 1 + \Big({dx \over dy}\Big)^2 } \ dy$

$= 4 \pi \int_{0}^{1} (1-y^{2/3})^{3/2} \sqrt{ 1 + \Big( {-(1-y^{2/3})^{1/2} \over y^{1/3}} \Big)^2 } \ dy$

$= 4 \pi \int_{0}^{1} (1-y^{2/3})^{3/2} \sqrt{ 1 + { {1-y^{2/3}} \over y^{2/3}} } \ dy$

$= 4 \pi \int_{0}^{1} (1-y^{2/3})^{3/2} \sqrt{ { {y^{2/3} + (1-y^{2/3}) } \over y^{2/3}} } \ dy$

$= 4 \pi \int_{0}^{1} (1-y^{2/3})^{3/2} \sqrt{ { 1 \over y^{2/3} } } \ dy$

$= 4 \pi \int_{0}^{1} (1-y^{2/3})^{3/2} \cdot { 1 \over y^{1/3} } \ dy$

$= 4 \pi \int_{0}^{1} { (1-y^{2/3})^{3/2} \over y^{1/3} } \ dy$

$\Big($ Let's do a

$u$ -substitution. Let

$u= 1-y^{2/3} \$ so that

$\ du = (-2/3)y^{-1/3} dy \ \ \longrightarrow \ \ (-3/2) du = {1 \over y^{1/3}} dy.\ \$ Since

$\ y:0 \rightarrow 1 \$ and

$\ u= 1-y^{2/3} \$ , it follows that

$\ u:1 \rightarrow 0. \$ Now make the substitutions.

$\Big)$

$= 4 \pi \int_{1}^{0} u^{3/2} \ (-3/2) du$

$= 4 \pi (-3/2) \int_{1}^{0} u^{3/2} \ du$

$= -6 \pi \cdot (2/5)u^{5/2} \Big|_{1}^{0}$

$= -6 \pi \Big( (2/5)(0)^{5/2} - (2/5)(1)^{5/2} \Big)$

$= -6 \pi ( 0 - 2/5 )$

$= (12/5) \pi$

Click HERE to return to the list of problems.