Finding the volume of regions rotated about the x - axis

[shamieh]

Need someone to verify that my work is correct please. Consider the region bounded by $y = sin(x)$ and the x - axis from $ x = 0$ to $x = \pi$

a) Find the volume if the region is rotated about the x - axis.

\(\displaystyle V = \int \pi (sin(x))^2 \, dx\)

\(\displaystyle \pi \int^{\pi}_0 sin^2x \, dx\)

\(\displaystyle \pi \int^{\pi}_0 \frac{(1 - cos2x)}{2} \, dx\)

\(\displaystyle \frac{1}{2} \pi \int^{\pi}_0 cos2x \, dx\)

\(\displaystyle \frac{1}{4} \pi \int^{\pi}_0 cos(u) \, dx\)

\(\displaystyle u = 2x\)
\(\displaystyle \frac{du}{2} = dx\)

After updating the limits I get $\int^0_0$ thus: $0$

So I'm left with \(\displaystyle \frac{1}{4}\pi\)

 

[shamieh]

b) Find the volume if this region is rotated about the y-axis

So for this one I got:

\(\displaystyle \frac{1}{4} \pi [sin(2x)] |^2_0 = \frac{sin(4)}{4\pi}\)

(sorry, meant to include this with my problem above, this was part b.)

 

[MarkFL]

For part a) I agree with you up to this point:

\(\displaystyle \pi\int^{\pi}_0 \frac{1-\cos(2x)}{2}\,dx\)

At this point I would bring the integrand's factor of \(\displaystyle \frac{1}{2}\) out front:

\(\displaystyle \frac{\pi}{2}\int^{\pi}_0 1-\cos(2x)\,dx\)

Now, use the substitution:

\(\displaystyle u=2x\,\therefore\,du=2\,dx\)

and rewrite the definite integral, making sure to change the limits in accordance with the substitution, that is, change them from $x$'s to $u$'s:

\(\displaystyle \frac{\pi}{4}\int^{2\pi}_0 1-\cos(u)\,du\)

Can you continue?

 

[MarkFL]

b) Find the volume if this region is rotated about the y-axis

So for this one I got:

\(\displaystyle \frac{1}{4} \pi [sin(2x)] |^2_0 = \frac{sin(4)}{4\pi}\)

(sorry, meant to include this with my problem above, this was part b.)

I would use the shell method here. The volume of an arbitrary shell is:

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

Can you identify the radius and height of the cylindrical shell in terms of $x$?

 

[shamieh]

For part a) I agree with you up to this point:

\(\displaystyle \pi\int^{\pi}_0 \frac{1-\cos(2x)}{2}\,dx\)

At this point I would bring the integrand's factor of \(\displaystyle \frac{1}{2}\) out front:

\(\displaystyle \frac{\pi}{2}\int^{\pi}_0 1-\cos(2x)\,dx\)

Now, use the substitution:

\(\displaystyle u=2x\,\therefore\,du=2\,dx\)

and rewrite the definite integral, making sure to change the limits in accordance with the substitution, that is, change them from $x$'s to $u$'s:

\(\displaystyle \frac{\pi}{4}\int^{2\pi}_0 1-\cos(u)\,du\)

Can you continue?

Oh I see what you're saying. I see what I did wrong now.

so you end up with \(\displaystyle \frac{1}{4} \pi [ -\sin(2x)] |^{2\pi}_0\)
And you know that \(\displaystyle \sin(2\pi)\) is just \(\displaystyle 0\) so \(\displaystyle \sin(4\pi)\) must be \(\displaystyle 0\) and \(\displaystyle \sin(2(0)) = 0\) so therefore \(\displaystyle V = \frac{1}{4} \pi\) ?

 

[MarkFL]

Oh I see what you're saying. I see what I did wrong now.

so you end up with \(\displaystyle \frac{1}{4} \pi [ -\sin(2x)] |^{2\pi}_0\)
And you know that \(\displaystyle \sin(2\pi)\) is just \(\displaystyle 0\) so \(\displaystyle \sin(4\pi)\) must be \(\displaystyle 0\) and \(\displaystyle \sin(2(0)) = 0\) so therefore \(\displaystyle V = \frac{1}{4} \pi\) ?

You are not using the correct anti-derivative in your application of the FTOC. What is:

\(\displaystyle \int 1-\cos(u)\,du\) ?

 

[shamieh]

You are not using the correct anti-derivative in your application of the FTOC. What is:

\(\displaystyle \int 1-\cos(u)\,du\) ?

Oh wow I'm an idiot. Is this right?

\(\displaystyle \frac{1}{4} \pi [ u - sin(u)]\)

\(\displaystyle \frac{1}{16}\pi\)

 

[MarkFL]

Oh wow I'm an idiot. Is this right?

\(\displaystyle \frac{1}{4} \pi [ u - sin(u)]\)

\(\displaystyle \frac{1}{16}\pi\)

You now have the correct anti-derivative. What you want is:

\(\displaystyle V=\frac{\pi}{4}\left[u-\sin(u) \right]_0^{2\pi}\)

Try the evaluation again with the limits. Recall that the notation means:

\(\displaystyle \left[f(u) \right]_a^b=f(b)-f(a)\)

 

[shamieh]

You now have the correct anti-derivative. What you want is:

\(\displaystyle V=\frac{\pi}{4}\left[u-\sin(u) \right]_0^{2\pi}\)

Try the evaluation again with the limits. Recall that the notation means:

\(\displaystyle \left[f(u) \right]_a^b=f(b)-f(a)\)

Wow I'm still an idiot lol multiplied the denominator by the denominator, maybe I should go back to high school?(Punch) \(\displaystyle \frac{4\pi}{4\pi} = 1\)

 

[MarkFL]

Wow I'm still an idiot lol multiplied the denominator by the denominator, maybe I should go back to high school?(Punch) \(\displaystyle \frac{4\pi}{4\pi} = 1\)

What you want to do here is:

\(\displaystyle V=\frac{\pi}{4}\left[u-\sin(u) \right]_0^{2\pi}=\frac{\pi}{4}\left(\left(2\pi-\sin(2\pi) \right)-\left(0-\sin(0) \right) \right)=\frac{\pi}{4}\left(2\pi \right)=\frac{\pi^2}{2}\)

 

[shamieh]

Should I set up the y-axis as so: \(\displaystyle V = \int ^1_0 2\pi(sinx) \, dx\)

 

[MarkFL]

Should I set up the y-axis as so: \(\displaystyle V = \int ^1_0 2\pi(sinx) \, dx\)

You are missing the radius of the shells and your upper limit is incorrect.

 

[shamieh]

What you want to do here is:

\(\displaystyle V=\frac{\pi}{4}\left[u-\sin(u) \right]_0^{2\pi}=\frac{\pi}{4}\left(\left(2\pi-\sin(2\pi) \right)-\left(0-\sin(0) \right) \right)=\frac{\pi}{4}\left(2\pi \right)=\frac{\pi^2}{2}\)

what? I thought you would unravel the u first making it \(\displaystyle \frac{1}{4}\pi [ 2x - sin(2x)] |^{2\pi}_0
\)

and then substitute your \(\displaystyle 2\pi\) and 0 in?

 

[MarkFL]

what? I thought you would unravel the u first making it \(\displaystyle \frac{1}{4}\pi [ 2x - sin(2x)] |^{2\pi}_0
\)

and then substitute your \(\displaystyle 2\pi\) and 0 in?

When you make a substitution in a definite integral, and rewrite everything (the integrand, the differential, and the limits) in terms of the new variable, you needn't worry about the old variable again...you are done with it for good. :D

 

[shamieh]

When you make a substitution in a definite integral, and rewrite everything (the integrand, the differential, and the limits) in terms of the new variable, you needn't worry about the old variable again...you are done with it for good. :D

Wow. Thanks so much for all the help Mark, sorry if I'm being a pain it just takes forever for me to grasp things and Calculus II is insane compared to last years Calc 1 class lol. I will be back on here soon. Going to try and go work with the y - axis problems and get my integral set up right. Thanks again for all the help.

Sham

 

[MarkFL]

Wow. Thanks so much for all the help Mark, sorry if I'm being a pain it just takes forever for me to grasp things and Calculus II is insane compared to last years Calc 1 class lol. I will be back on here soon. Going to try and go work with the y - axis problems and get my integral set up right. Thanks again for all the help.

Sham

I am happy to help, and you are making an effort to learn and are not a "pain" at all. :D

 

[shamieh]

You are missing the radius of the shells and your upper limit is incorrect.

I think I finally figured it out! So x is the distance to my actual shell I cut out so then taking top function minus bottom

\(\displaystyle \int 2\pi\) (shell radius)(shell height)

which means I will have

\(\displaystyle 2\pi \int^{\pi}_0 (x)(\sin x) \,dx \) is that right?

 

[MarkFL]

I think I finally figured it out! So x is the distance to my actual shell I cut out so then taking top function minus bottom

\(\displaystyle \int 2\pi\) (shell radius)(shell height)

which means I will have

\(\displaystyle 2\pi \int^{\pi}_0 (x)(\sin x) \,dx \) is that right?

Yes, you have the correct definite integral representing the volume of the described solid of revolution. :D

Can you set up an integral representing the volume using the washer method?

 

[shamieh]

so would this be correct for question b)?

This is the solution I got\(\displaystyle 2\pi \int^{\pi}_0 x sinx \, dx\)

\(\displaystyle u = x\)
\(\displaystyle du = dx\)

\(\displaystyle dv = cosx\)
\(\displaystyle v = sinx\)

\(\displaystyle 2\pi [[xsinx - \int sinx \,dx\)

\(\displaystyle 2\pi[[xsinx - cosx |^{\pi}_0 = 4\pi\)

 

[MarkFL]

so would this be correct for question b)?

This is the solution I got\(\displaystyle 2\pi \int^{\pi}_0 x sinx \, dx\)

\(\displaystyle u = x\)
\(\displaystyle du = dx\)

\(\displaystyle dv = cosx\)
\(\displaystyle v = sinx\)

\(\displaystyle 2\pi [[xsinx - \int sinx \,dx\)

\(\displaystyle 2\pi[[xsinx - cosx |^{\pi}_0 = 4\pi\)

You have set up your integration by parts incorrectly. You want to use:

\(\displaystyle \int_a^b u\,dv=\left[uv \right]_a^b-\int_a^v v\,du\)

You original integrand and differential is:

\(\displaystyle x\sin(x)\,dx\)

By choosing:

\(\displaystyle u=x\,\therefore\,du=dx\)

You need to the let $dv$ be everything that is left:

\(\displaystyle dv=\sin(x)\,dx\,\therefore\,v=-\cos(x)\)

So you know have:

\(\displaystyle 2\pi\int_0^{\pi}x\sin(x)\,dx=2\pi\left(\left[-x\cos(x) \right]_0^{\pi}+\int_0^{\pi}\cos(x)\,dx \right)\)

What do you then find?

 

[shamieh]

You have set up your integration by parts incorrectly. You want to use:

\(\displaystyle \int_a^b u\,dv=\left[uv \right]_a^b-\int_a^v v\,du\)

You original integrand and differential is:

\(\displaystyle x\sin(x)\,dx\)

By choosing:

\(\displaystyle u=x\,\therefore\,du=dx\)

You need to the let $dv$ be everything that is left:

\(\displaystyle dv=\sin(x)\,dx\,\therefore\,v=-\cos(x)\)

So you know have:

\(\displaystyle 2\pi\int_0^{\pi}x\sin(x)\,dx=2\pi\left(\left[-x\cos(x) \right]_0^{\pi}+\int_0^{\pi}\cos(x)\,dx \right)\)

What do you then find?

Oh I see. \(\displaystyle 2\pi\) is what you find correct?

 

[MarkFL]

Oh I see. \(\displaystyle 2\pi\) is what you find correct?

Not quite. Can you show your work so I can see what you did?

 

[shamieh]

\(\displaystyle 2\pi \int^\pi_0 (x)(sinx)dx
\)
\(\displaystyle u = x\)
\(\displaystyle du = dx\)
\(\displaystyle dv = sinxdx\)
\(\displaystyle v = -cosx\)

\(\displaystyle uv - \int vdu\)

\(\displaystyle -xcosx + \int cosxdx\)

\(\displaystyle 2\pi[-xcosx + sinx] |^{\pi}_0 = 2\pi\)

 

[MarkFL]

How did you get

\(\displaystyle 2\pi\left[-x\cos(x)+\sin(x) \right]_0^{\pi}=2\pi\) ?

 

[shamieh]

Wow I'm the stupidest human being on planet earth. I was saying cos(pi) = -1 and forgetting about the other x infront of the sin

so \(\displaystyle -\pi * -1 = \pi\)

so \(\displaystyle 2\pi ([ \pi ] - [0] = 2 \pi ^2\)

 

[MarkFL]

Wow I'm the stupidest human being on planet earth. I was saying cos(pi) = -1 and forgetting about the other x infront of the sin

so \(\displaystyle -\pi * -1 = \pi\)

so \(\displaystyle 2\pi ([ \pi ] - [0] = 2 \pi ^2\)

Yes, that is correct! (Cool)

Can you work this problem using the washer method? (Muscle)

 

[shamieh]

Yes, that is correct! (Cool)

Can you work this problem using the washer method? (Muscle)

I'm kind of confused on the washer method because isn't it \(\displaystyle \int \pi (Area Of Outer)^2 - \pi(AreaOfInner)^2 dx\)? So the question is what is my outer function..But I'm bounded at \(\displaystyle \pi\) so Would it be \(\displaystyle \int \pi(x)^2 - Area Of Outer ^2\) <-- which I'm not sure what this would be because there is no outer? Again I'm just kind of guessing.

 

[MarkFL]

Your outer radius would be the distance from the portion of the curve on \(\displaystyle \left[\frac{\pi}{2},\pi \right]\) to the line $x=0$ and the inner radius would be the distance from the portion of the curve on \(\displaystyle \left[0,\frac{\pi}{2} \right]\) to the line $x=0$. Draw a sketch and you will see what I'm talking about.

Now, you will want these radii in terms of $y$ since the thickness of each washer is $dy$. So, you will want to make use of the following identity for the outer radius:

\(\displaystyle \sin(\pi-\theta)=\sin(\theta)\)

So, for the inner radius $r$, we begin with:

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

Hence:

\(\displaystyle r=\sin^{-1}(y)-0=\sin^{-1}(y)\)

And for the outer radius $R$, we begin with:

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

Hence:

\(\displaystyle R=\pi-\sin^{-1}(y)\)

So, you want to use for the volume of an arbitrary washer:

\(\displaystyle dV=\pi\left(R^2-r^2 \right)\,dy\)

Can you now construct the integral? :D