Help with volume of revolution question

[student93]

Homework Statement

Find the volume obtained by rotating the region between the graph of y=0.5(sin(x^2)^2) and the x-axis (from 0 to squareroot pi) about the y-axis.

The answer pi^2/4, but I don't understand how to get the answer, I can set up the integral but can't simplify it to that answer.

Homework Equations

y=0.5(sin(x^2)^2)

The Attempt at a Solution

First to get the radius in terms of y -> y=0.5sin^2(x^2), thus √arcsin(√2y)=x

The volume is thus calculated via the disk method and is equal to the integral of pi√arcsin(√2y), from 0 to √pi. I don't know where to go form there,maybe I didn't set up the integral correctly?

 

[Saitama]

Homework Statement

Find the volume obtained by rotating the region between the graph of y=0.5(sin(x^2)^2) and the x-axis (from 0 to squareroot pi) about the y-axis.

The answer pi^2/4, but I don't understand how to get the answer, I can set up the integral but can't simplify it to that answer.

Homework Equations

y=0.5(sin(x^2)^2)

The Attempt at a Solution

First to get the radius in terms of y -> y=0.5sin^2(x^2), thus √arcsin(√2y)=x

The volume is thus calculated via the disk method and is equal to the integral of pi√arcsin(√2y), from 0 to √pi. I don't know where to go form there,maybe I didn't set up the integral correctly?

Hi thisisssu!

I suggest using method of cylinders instead of disk as we are rotating about the y-axis.

 

[vela]

Homework Statement

Find the volume obtained by rotating the region between the graph of y=0.5(sin(x^2)^2) and the x-axis (from 0 to squareroot pi) about the y-axis.

The answer pi^2/4, but I don't understand how to get the answer, I can set up the integral but can't simplify it to that answer.

Homework Equations

y=0.5(sin(x^2)^2)

The Attempt at a Solution

First to get the radius in terms of y -> y=0.5sin^2(x^2), thus √arcsin(√2y)=x

The radius of what?

The volume is thus calculated via the disk method and is equal to the integral of pi√arcsin(√2y), from 0 to √pi. I don't know where to go form there,maybe I didn't set up the integral correctly?

I think you're saying that
$$V = \int_0^\sqrt\pi \pi r\,dy,$$ where r is the radius you describe above. This isn't a correct application of the disk method. Also, I'm assuming you're trying to integrate with respect to y because you wrote the integrand in terms of y. If that's the case, your limits aren't correct because those are x values.

In any case, the disk method is not a good choice for this particular problem. Take Pranav-Arora's advice and use cylinders.

 

[student93]

I've attached the problem in this post to avoid any confusion, also by "cylinder method", I'm assuming you're actually referring to the shells method?

 

[Saitama]

I've attached the problem in this post to avoid any confusion, also by "cylinder method", I'm assuming you're actually referring to the shells method?

Yes!

 

[student93]

So I tried the shell method and ended up getting pi/2 which is not the correct answer. I was able to set up the integral which was the integral of 2pi0.5sin^2(x^2)xdx from 0 to √pi etc., with the height and radius being 0.5sin^2(x^2) and x respectively. Is my setup wrong?

 

[vela]

No, that looks right. It seems you didn't evaluate the integral correctly.

 

[student93]

I verified via wolframaalpha, and it doesn't come out to pi^2/4, but 1.56 which is essentially pi/2, so I'm assuming I didn't set up integral correctly or maybe I did screw up somewhere while trying to solve the integral? Could someone help me evaluate this integral.

 

[vela]

You're typing it in wrong to Wolfram Alpha then. Show us your work so we can point out where you're going astray.

 

[student93]

V=2π∫(1/2)(sin^2(x^2))(x)dx, from 0 to √π
V=π/2∫sin^2(u)du, from 0 to √π, with u=x^2 and du=2xdx (I started having trouble simplifying after getting up to here).

 

[vela]

That's almost correct. You should have
$$ V = \frac{\pi}{2} \int_0^\pi \sin^2 u\,du.$$ The upper limit is $\pi$, not $\sqrt\pi$. You want to use a trig identity for $\sin^2 u$.

 

[student93]

That's almost correct. You should have
$$ V = \frac{\pi}{2} \int_0^\pi \sin^2 u\,du.$$ The upper limit is $\pi$, not $\sqrt\pi$. You want to use a trig identity for $\sin^2 u$.

Could you please explain why the upper limit is π? Also how did you calculate the upper limit as π?

 

[haruspex]

Could you please explain why the upper limit is π? Also how did you calculate the upper limit as π?

The limits are x=0 to x = √π. (Without the "x=" they're incomplete, technically.) With u = x², that's u=x²=0 to u=x²=π.

 

[student93]

Thanks, I finally ended up getting the correct answer after using the correct limits.