find the area of surface generated by revolving the curve about y -axis

Mar 2014
909
2
malaysia
i am stucked here , how to continue ? DSC_0330[1].JPG
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
Hey xl5899.

I have noticed that you are pulling a non-constant term out of the integral. Can you please explain why this is?
 
Mar 2014
909
2
malaysia
Hey xl5899.

I have noticed that you are pulling a non-constant term out of the integral. Can you please explain why this is?
i will put it into the integral later , but my problem is how to bring the y^-1/3 inside ? since there is sqrt rt there...
 

Prove It

MHF Helper
Aug 2008
12,897
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The point is that you can not EVER pull out the variable you are integrating with respect to, regardless of whether you are planning to put it back! You are not going to get very far if you continue to do what you please instead of what is mathematically correct!

To answer the rest of your question, when the working is fixed up...

$\displaystyle \begin{align*} 2\,\pi \int_{-1}^0{y^{\frac{1}{3}} \, \frac{\sqrt{9\,y^{\frac{4}{3}} + 1}}{3\,y^{\frac{2}{3}}} \,\mathrm{d}y} &= \frac{\pi}{18} \int_{-1}^0{ 12\,y^{\frac{1}{3}}\,\frac{\sqrt{9\,y^{\frac{4}{3}} + 1}}{y^{\frac{2}{3}}}\,\mathrm{d}y } \end{align*}$

Now let $\displaystyle \begin{align*} u = 9\,y^{\frac{4}{3}} + 1 \implies \mathrm{d}u = 12\,y^{\frac{1}{3}}\,\mathrm{d}y \end{align*}$, noting that $\displaystyle \begin{align*} u\left( -1 \right) = 9 \end{align*}$ and $\displaystyle \begin{align*} u \left( 0 \right) = 0 \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \frac{\pi}{18} \int_{-1}^0{ 12\,y^{\frac{1}{3}}\,\frac{\sqrt{9\,y^{\frac{4}{3}} + 1}}{y^{\frac{2}{3}}} \,\mathrm{d}y } &= \frac{\pi}{18} \int_{9}^0{ \frac{\sqrt{u}}{\frac{\sqrt{u-1}}{3} } \,\mathrm{d}u } \\ &= -\frac{\pi}{6} \int_0^9{ \frac{\sqrt{u}}{\sqrt{u - 1}} \,\mathrm{d}u } \\ &= -\frac{\pi}{6} \int_0^9{ \frac{u}{\sqrt{u}\,\sqrt{u - 1}} \,\mathrm{d}u } \\ &= \frac{\pi}{3} \int_0^9{ u\,\left( -\frac{1}{2\,\sqrt{u}\,\sqrt{u - 1} } \right) \,\mathrm{d}u } \end{align*}$

Now let $\displaystyle \begin{align*} v = \sqrt{ \frac{u}{u - 1} } \implies \mathrm{d}v = -\frac{1}{2\,\sqrt{u}\,\sqrt{ u - 1} } \,\mathrm{d}u \end{align*}$, note that $\displaystyle \begin{align*} v \left( 0 \right) = 0 \end{align*}$ and $\displaystyle \begin{align*} v \left( 9 \right) = \sqrt{ \frac{9}{8} } = \frac{3}{2\,\sqrt{2}} = \frac{3\,\sqrt{2}}{4} \end{align*}$. Also make sure you can evaluate u in terms of v:

$\displaystyle \begin{align*} v &= \sqrt{ \frac{u}{u-1} } \\ v^2 &= \frac{u}{u - 1} \\ v^2 \,\left( u - 1 \right) &= u \\ v^2\,u - v^2 &= u \\ v^2\,u - u &= v^2 \\ u\,\left( v^2 - 1 \right) &= v^2 \\ u &= \frac{v^2}{v^2 - 1} \end{align*}$

and the integral becomes

$\displaystyle \begin{align*} \frac{\pi}{3} \int_0^9{ u\,\left( -\frac{1}{2\,\sqrt{u}\,\sqrt{u-1}} \right) \,\mathrm{d}u } &= \frac{\pi}{3} \int_0^{\frac{3\,\sqrt{2}}{4}}{ \frac{v^2}{v^2 - 1}\,\mathrm{d}v } \\ &= \frac{\pi}{3} \int_0^{\frac{3\,\sqrt{2}}{4}}{ \left( 1 + \frac{1}{v^2 - 1} \right) \,\mathrm{d}v } \\ &= \frac{\pi}{3} \int_0^{\frac{3\,\sqrt{2}}{4}}{ \left[ 1 + \frac{1}{ \left( v - 1 \right) \left( v + 1 \right) } \right] \,\mathrm{d}v } \\ &= \frac{\pi}{3} \int_0^{\frac{3\,\sqrt{2}}{4}}{ \left( 1 + \frac{\frac{1}{2}}{v - 1} - \frac{\frac{1}{2}}{v + 1} \right) \,\mathrm{d}v } \\ &= \frac{\pi}{3} \,\left[ v + \frac{1}{2} \ln{ \left| v - 1 \right| } - \frac{1}{2} \ln{ \left| v + 1 \right| } \right] _0^{ \frac{3\,\sqrt{2}}{4} } \\ &= \frac{\pi}{3} \, \left[ v + \frac{1}{2} \ln{ \left| \frac{v - 1}{v + 1} \right| } \right] _0^{\frac{3\,\sqrt{2}}{4}} \\ &= \frac{\pi}{3} \, \left[ \left( \frac{3\,\sqrt{2}}{4} + \frac{1}{2} \ln{ \left| \frac{\frac{3\,\sqrt{2}}{4} - 1}{\frac{3\,\sqrt{2}}{4} + 1} \right| } \right) - \left( 0 + \frac{1}{2} \ln{ \left| \frac{0 - 1}{0 + 1} \right| } \right) \right] \\ &= \frac{\pi}{3} \, \left[ \frac{3\,\sqrt{2}}{4} + \frac{1}{2} \ln{ \left( \frac{ 3\,\sqrt{2} - 4 }{ 3\,\sqrt{2} + 4 } \right) } - \frac{1}{2} \ln{ \left| -1 \right| } \right] \\ &= \frac{\pi}{3} \left[ \frac{3\,\sqrt{2}}{4} + \frac{1}{2} \ln{ \left( \frac{3\,\sqrt{2} - 4}{3\,\sqrt{2} + 4} \right) } \right] \\ &= \frac{\pi\,\left[ 3\,\sqrt{2} + 2 \ln{ \left( 3\,\sqrt{2} - 4 \right) } - 2 \ln{ \left( 3\,\sqrt{2} + 4 \right) } \right] }{12} \end{align*}$
 
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Mar 2014
909
2
malaysia
The point is that you can not EVER pull out the variable you are integrating with respect to, regardless of whether you are planning to put it back! You are not going to get very far if you continue to do what you please instead of what is mathematically correDSC_0333[1].JPGct!

To answer the rest of your question, when the working is fixed up...

$\displaystyle \begin{align*} 2\,\pi \int_{-1}^0{y^{\frac{1}{3}} \, \frac{\sqrt{9\,y^{\frac{4}{3}} + 1}}{3\,y^{\frac{2}{3}}} \,\mathrm{d}y} &= \frac{\pi}{18} \int_{-1}^0{ 12\,y^{\frac{1}{3}}\,\frac{\sqrt{9\,y^{\frac{4}{3}} + 1}}{y^{\frac{2}{3}}}\,\mathrm{d}y } \end{align*}$

Now let $\displaystyle \begin{align*} u = 9\,y^{\frac{4}{3}} + 1 \implies \mathrm{d}u = 12\,y^{\frac{1}{3}}\,\mathrm{d}y \end{align*}$, noting that $\displaystyle \begin{align*} u\left( -1 \right) = 9 \end{align*}$ and $\displaystyle \begin{align*} u \left( 0 \right) = 0 \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \frac{\pi}{18} \int_{-1}^0{ 12\,y^{\frac{1}{3}}\,\frac{\sqrt{9\,y^{\frac{4}{3}} + 1}}{y^{\frac{2}{3}}} \,\mathrm{d}y } &= \frac{\pi}{18} \int_{9}^0{ \frac{\sqrt{u}}{\frac{\sqrt{u-1}}{3} } \,\mathrm{d}u } \\ &= -\frac{\pi}{6} \int_0^9{ \frac{\sqrt{u}}{\sqrt{u - 1}} \,\mathrm{d}u } \\ &= -\frac{\pi}{6} \int_0^9{ \frac{u}{\sqrt{u}\,\sqrt{u - 1}} \,\mathrm{d}u } \\ &= \frac{\pi}{3} \int_0^9{ u\,\left( -\frac{1}{2\,\sqrt{u}\,\sqrt{u - 1} } \right) \,\mathrm{d}u } \end{align*}$

Now let $\displaystyle \begin{align*} v = \sqrt{ \frac{u}{u - 1} } \implies \mathrm{d}v = -\frac{1}{2\,\sqrt{u}\,\sqrt{ u - 1} } \,\mathrm{d}u \end{align*}$, note that $\displaystyle \begin{align*} v \left( 0 \right) = 0 \end{align*}$ and $\displaystyle \begin{align*} v \left( 9 \right) = \sqrt{ \frac{9}{8} } = \frac{3}{2\,\sqrt{2}} = \frac{3\,\sqrt{2}}{4} \end{align*}$. Also make sure you can evaluate u in terms of v:

$\displaystyle \begin{align*} v &= \sqrt{ \frac{u}{u-1} } \\ v^2 &= \frac{u}{u - 1} \\ v^2 \,\left( u - 1 \right) &= u \\ v^2\,u - v^2 &= u \\ v^2\,u - u &= v^2 \\ u\,\left( v^2 - 1 \right) &= v^2 \\ u &= \frac{v^2}{v^2 - 1} \end{align*}$

and the integral becomes

$\displaystyle \begin{align*} \frac{\pi}{3} \int_0^9{ u\,\left( -\frac{1}{2\,\sqrt{u}\,\sqrt{u-1}} \right) \,\mathrm{d}u } &= \frac{\pi}{3} \int_0^{\frac{3\,\sqrt{2}}{4}}{ \frac{v^2}{v^2 - 1}\,\mathrm{d}v } \\ &= \frac{\pi}{3} \int_0^{\frac{3\,\sqrt{2}}{4}}{ \left( 1 + \frac{1}{v^2 - 1} \right) \,\mathrm{d}v } \\ &= \frac{\pi}{3} \int_0^{\frac{3\,\sqrt{2}}{4}}{ \left[ 1 + \frac{1}{ \left( v - 1 \right) \left( v + 1 \right) } \right] \,\mathrm{d}v } \\ &= \frac{\pi}{3} \int_0^{\frac{3\,\sqrt{2}}{4}}{ \left( 1 + \frac{\frac{1}{2}}{v - 1} - \frac{\frac{1}{2}}{v + 1} \right) \,\mathrm{d}v } \\ &= \frac{\pi}{3} \,\left[ v + \frac{1}{2} \ln{ \left| v - 1 \right| } - \frac{1}{2} \ln{ \left| v + 1 \right| } \right] _0^{ \frac{3\,\sqrt{2}}{4} } \\ &= \frac{\pi}{3} \, \left[ v + \frac{1}{2} \ln{ \left| \frac{v - 1}{v + 1} \right| } \right] _0^{\frac{3\,\sqrt{2}}{4}} \\ &= \frac{\pi}{3} \, \left[ \left( \frac{3\,\sqrt{2}}{4} + \frac{1}{2} \ln{ \left| \frac{\frac{3\,\sqrt{2}}{4} - 1}{\frac{3\,\sqrt{2}}{4} + 1} \right| } \right) - \left( 0 + \frac{1}{2} \ln{ \left| \frac{0 - 1}{0 + 1} \right| } \right) \right] \\ &= \frac{\pi}{3} \, \left[ \frac{3\,\sqrt{2}}{4} + \frac{1}{2} \ln{ \left( \frac{ 3\,\sqrt{2} - 4 }{ 3\,\sqrt{2} + 4 } \right) } - \frac{1}{2} \ln{ \left| -1 \right| } \right] \\ &= \frac{\pi}{3} \left[ \frac{3\,\sqrt{2}}{4} + \frac{1}{2} \ln{ \left( \frac{3\,\sqrt{2} - 4}{3\,\sqrt{2} + 4} \right) } \right] \\ &= \frac{\pi\,\left[ 3\,\sqrt{2} + 2 \ln{ \left( 3\,\sqrt{2} - 4 \right) } - 2 \ln{ \left( 3\,\sqrt{2} + 4 \right) } \right] }{12} \end{align*}$
your method is rather complicated , how if i wanna do in this way ? but , i dunno how to continue form here. Can you explain ?
 

Prove It

MHF Helper
Aug 2008
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You don't know how to continue? Try reading the response, considering I ANSWERED THE WHOLE DAMN QUESTION!!!