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

::<math>y^3=x</math>

between  <math style="vertical-align: -5px">(0,0)</math>  and  <math style="vertical-align: -5px">(1,1)</math>  about the  <math style="vertical-align: -4px">y</math>-axis.

(b) Find the length of the arc

::<math>y=1+9x^{\frac{3}{2}}</math>

between the points  <math style="vertical-align: -5px">(1,10)</math>  and  <math style="vertical-align: -5px">(4,73).</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' The surface area  <math style="vertical-align: 0px">S</math>  of a function  <math style="vertical-align: -5px">y=f(x)</math>  rotated about the  <math style="vertical-align: -4px">y</math>-axis is given by
|-
|
       <math style="vertical-align: -13px">S=\int 2\pi x\,ds,</math>  where  <math style="vertical-align: -19px">ds=\sqrt{1+\bigg(\frac{dx}{dy}\bigg)^2}dy.</math>
|-
|'''2.''' The formula for the length  <math style="vertical-align: 0px">L</math>  of a curve  <math style="vertical-align: -5px">y=f(x)</math>  where  <math style="vertical-align: -3px">a\leq x \leq b</math>  is
|-
|
       <math>L=\int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We start by calculating  <math style="vertical-align: -16px">\frac{dx}{dy}.</math>
|-
|Since  <math style="vertical-align: -4px">x=y^3,</math>
|-
|       <math>\frac{dx}{dy}=3y^2.</math>
|-
|Now, we are going to integrate with respect to  <math style="vertical-align: -3px">y.</math>
|-
|Using the formula given in the Foundations section,
|-
|we have
|-
|        <math>\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}</math>
|-
|where  <math style="vertical-align: 0px">S</math>  is the surface area.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we use  <math style="vertical-align: 0px">u</math>-substitution.
|-
|Let  <math style="vertical-align: -5px">u=1+9y^4.</math>
|-
|Then,  <math style="vertical-align: -5px">du=36y^3dy</math>  and  <math style="vertical-align: -14px">\frac{du}{36}=y^3dy.</math>
|-
|Also, since this is a definite integral, we need to change the bounds of integration.
|-
|We have
|-
|       <math style="vertical-align: -5px">u_1=1+9(0)^4=1</math>  and  <math style="vertical-align: -5px">u_2=1+9(1)^4=10.</math>
|-
|Thus, we get
|-
|        <math>\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}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we calculate  <math style="vertical-align: -15px">\frac{dy}{dx}.</math>
|-
|Since  <math style="vertical-align: -5px">y=1+9x^{\frac{3}{2}},</math>  we have
|-
|       <math>\frac{dy}{dx}=\frac{27\sqrt{x}}{2}.</math>
|-
|Then, the arc length  <math style="vertical-align: 0px">L</math>  of the curve is given by
|-
|       <math>L=\int_1^4 \sqrt{1+\bigg(\frac{27\sqrt{x}}{2}\bigg)^2}~dx.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Then, we have
|-
|       <math>L=\int_1^4 \sqrt{1+\frac{27^2x}{2^2}}~dx.</math>
|-
|Now, we use  <math style="vertical-align: 0px">u</math>-substitution.
|-
|Let  <math style="vertical-align: -14px">u=1+\frac{27^2x}{2^2}.</math>
|-
|Then,  <math style="vertical-align: -14px">du=\frac{27^2}{2^2}dx</math>  and  <math style="vertical-align: -14px">dx=\frac{2^2}{27^2}du.</math>
|-
|Also, since this is a definite integral, we need to change the bounds of integration.
|-
|We have
|-
|       <math style="vertical-align: -13px">u_1=1+\frac{27^2(1)}{2^2}=1+\frac{27^2}{2^2}</math>  and  <math style="vertical-align: -13px">u_2=1+\frac{27^2(4)}{2^2}=1+27^2.</math>
|-
|Hence, we now have
|-
|       <math>L=\int_{1+\frac{27^2}{2^2}}^{1+27^2} \frac{2^2}{27^2}u^{\frac{1}{2}}~du.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Therefore, we have
|-
|        <math>\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}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>\frac{\pi}{27} (10)^{\frac{3}{2}}-\frac{\pi}{27}</math>
|-
|   '''(b)'''    <math>\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}}</math>
|}
[[009B_Sample_Final_2|'''Return to Sample Exam''']]

@@ Line 101: / Line 101: @@
 |Then, we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_1^4 \sqrt{1+\frac{27^2x}{2^2}}~dx.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_1^4 \sqrt{1+\frac{729x}{4}}~dx.</math>
 |-
 |Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|Let &nbsp;<math style="vertical-align: -14px">u=1+\frac{27^2x}{2^2}.</math>
+|Let &nbsp;<math style="vertical-align: -14px">u=1+\frac{729x}{4}.</math>
 |-
-|Then, &nbsp;<math style="vertical-align: -14px">du=\frac{27^2}{2^2}dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">dx=\frac{2^2}{27^2}du.</math>
+|Then, &nbsp;<math style="vertical-align: -14px">du=\frac{729}{4}~dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">dx=\frac{4}{729}~du.</math>
 |-
 |Also, since this is a definite integral, we need to change the bounds of integration.
@@ Line 113: / Line 113: @@
 |We have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">u_1=1+\frac{27^2(1)}{2^2}=1+\frac{27^2}{2^2}</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">u_2=1+\frac{27^2(4)}{2^2}=1+27^2.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">u_1=1+\frac{729(1)}{4}=\frac{733}{4}</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">u_2=1+\frac{729(4)}{4}=730.</math>
 |-
 |Hence, we now have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_{1+\frac{27^2}{2^2}}^{1+27^2} \frac{2^2}{27^2}u^{\frac{1}{2}}~du.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_{\frac{733}{4}}^{730} \frac{4}{729}u^{\frac{1}{2}}~du.</math>
 |}
@@ Line 126: / Line 126: @@
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\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{L} & = & \displaystyle{\frac{4}{729} \bigg(\frac{2}{3}u^{\frac{3}{2}}\bigg)\bigg|_{\frac{733}{4}}^{730}}\\
 &&\\
-& = & \displaystyle{\frac{2^3}{3^4} u^{\frac{3}{2}}\bigg|_{1+\frac{27^2}{2^2}}^{1+27^2}}\\
+& = & \displaystyle{\frac{8}{2187} u^{\frac{3}{2}}\bigg|_{\frac{733}{4}}^{730}}\\
 &&\\
-& = & \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}}.}
+& = & \displaystyle{\frac{8}{2187} (730)^{\frac{3}{2}}-\frac{8}{2187} \bigg(\frac{733}{4}\bigg)^{\frac{3}{2}}.}
 \end{array}</math>
 |}
@@ Line 140: / Line 140: @@
 |&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math>\frac{\pi}{27} (10)^{\frac{3}{2}}-\frac{\pi}{27}</math>
 |-
-|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>\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}}</math>
+|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>\frac{8}{2187} (730)^{\frac{3}{2}}-\frac{8}{2187} \bigg(\frac{733}{4}\bigg)^{\frac{3}{2}}</math>
 |}
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

009B Sample Final 2, Problem 5 - Revision history

MathAdmin at 01:08, 21 May 2017