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Shell method

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
Consider the region bounded by the lines \(\displaystyle x=0\),\(\displaystyle y=1\) and curve
\(\displaystyle y=\sin(x)\), \(\displaystyle 0\leq x \leq \frac{\pi}{2}\) decide the volume when it rotate in y-axe ( Tips: \(\displaystyle t=\sin^{-1}(y)\) can be usefull)
I got pretty confused what my integral limit shall be. I know the formula will be
\(\displaystyle \pi\int_a^b xf(x)^2 \).

Regards,
\(\displaystyle |\pi\rangle\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: shell method

The formula you have cited pertains to the disk method. After having sketched the region to be rotated, I find the volume of an arbitrary shell is:

\(\displaystyle dV=2\pi rh\,dx\)

where:

\(\displaystyle r=x\)

\(\displaystyle h=1-\sin(x)\)

However, given the hint you have included, I suspect you are to use the disk method instead. What is the volume of an arbitrary disk?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: shell method

The formula you have cited pertains to the disk method. After having sketched the region to be rotated, I find the volume of an arbitrary shell is:

\(\displaystyle dV=2\pi rh\,dx\)

where:

\(\displaystyle r=x\)

\(\displaystyle h=1-\sin(x)\)

However, given the hint you have included, I suspect you are to use the disk method instead. What is the volume of an arbitrary disk?
Hello Mark,
\(\displaystyle \int_a^bf(x)^2 = \int_a^b\arcsin^2(y)\) I am still kinda confused what our limit shall be
Regards,
\(\displaystyle |\pi\rangle\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: shell method

You have left off the differential from your definite integrals, which is the thickness of the disks, and will tell you with respect to which variable you are integrating, and thus, what your limits need to be. The volume of an arbitrary disk is:

\(\displaystyle dV=\pi r^2\,dy\)

where:

\(\displaystyle r=x=\sin^{-1}(y)\)

hence:

\(\displaystyle dV=\pi\left(\sin^{-1}(y) \right)^2\,dy\)

So, what values does $y$ have in the summation of the disks?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: shell method

You have left off the differential from you definite integrals, which is the thickness of the disks, and will tell you with respect to which variable you are integrating, and thus, what your limits need to be. The volume of an arbitrary disk is:

\(\displaystyle dV=\pi r^2\,dy\)

where:

\(\displaystyle r=x=\sin^{-1}(y)\)

hence:

\(\displaystyle dV=\pi\left(\sin^{-1}(y) \right)^2\,dy\)

So, what values does $y$ have in the summation of the disks?
Hello Mark,
If I have understand correctly, to get our y limit we got our x limit which is \(\displaystyle 0\leq x\leq\frac{\pi}{2}\) so our y limit will be \(\displaystyle \sin(0)=0\) and \(\displaystyle sin(\frac{\pi}{2})=1\) that means \(\displaystyle 0\leq y\leq1\) and about forgeting "dy" is something really bad cause I always think that in my and never write it up...

Regards,
\(\displaystyle |\pi\rangle\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: shell method

Yes, those are the correct limits of integration:

\(\displaystyle V=\pi\int_0^1\left(\sin^{-1}(y) \right)^2\,dy\)

How do you think you should proceed?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: shell method

Yes, those are the correct limits of integration:

\(\displaystyle V=\pi\int_0^1\left(\sin^{-1}(y) \right)^2\,dy\)

How do you think you should proceed?
Hello Mark,
I would do as they gave the tips. \(\displaystyle t=\sin^{-1}(y) \),\(\displaystyle \sin(t)=y <=> dy=\cos(t)dt\) for the limit \(\displaystyle y=0 <=> t=0\) and \(\displaystyle y=1 <=> t=\frac{\pi}{2}\) so we got
\(\displaystyle \pi\int_0^{\frac{\pi}{2}}t^2\cos(t)dt\)
is this correct?

Regards,
\(\displaystyle |\pi\rangle\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: shell method

Yes, that is correct! (Yes)
 

Petrus

Well-known member
Feb 21, 2013
739
Re: shell method

Yes, that is correct! (Yes)
Hello Mark,
First I wanna thank you for taking your time and helping me!:) I have correctly integrate ( for those who is interested you will have to use integration by part twice) but I am intresting on the method you posted #2 I don't se how you get \(\displaystyle h=1-sin(x)\) the answer is \(\displaystyle \frac{\pi(\pi^2-8)}{4}\) I would like to solve this problem with disc method as well.

Regards,
\(\displaystyle |\pi\rangle\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: shell method

Using the shell method, and referring to your sketch of the region to be revolved about the $y$=axis, can you see that the height of an arbitrary shell is the distance between the upper bound $y=1$ and the lower bound $y=\sin(x)$, and since on the given interval we have $\sin(x)\le1$, this distance is $1-\sin(x)$.
 

Petrus

Well-known member
Feb 21, 2013
739
Re: shell method

Using the shell method, and referring to your sketch of the region to be revolved about the $y$=axis, can you see that the height of an arbitrary shell is the distance between the upper bound $y=1$ and the lower bound $y=\sin(x)$, and since on the given interval we have $\sin(x)\le1$, this distance is $1-\sin(x)$.
Hello Mark,
this lead me to another question so basicly on what we did early our upper bound is \(\displaystyle \sin^{-1}(y)\) and our lower bound is \(\displaystyle y=0\) right? I think I did not understand that early but is this correct?

Regards,
\(\displaystyle |\pi\rangle\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: shell method

The bounds for $x$ are actually stated in the problem:

\(\displaystyle 0\le x\le\frac{\pi}{2}\)