Why do I get a negative value in solids of revolution?

[imperatormk]

So I have a circle with the following parametric equations:

. . . . .$\displaystyle \begin{cases}x&=&6\, +\, 2\cos(t)\\y&=&3\, +\, 2\sin(t)\end{cases}$

I have found that the lower limit is 0, and the upper 2pi. (Not sure if this is correct!)

Using the formula, I got the following integral:

. . . . .$\displaystyle \displaystyle \int_0^{6.28}\, \left(3\, +\, 2\sin(x)\right)^2\, \left(-2\sin(x)\right)\, dx$

(with pi before it, of course).

I get the correct value (48pi^2), however, it's negative.

I'm wondering if it's meant to be such or I have an error. I have even tried using an online calculator for the integral, and it also gives negative value for the integral.

Let me know if you need anything else.

Thanks!

 

[stapel]

So I have a circle with the following parametric equations:

. . . . .$\displaystyle \begin{cases}x&=&6\, +\, 2\cos(t)\\y&=&3\, +\, 2\sin(t)\end{cases}$

I have found that the lower limit is 0, and the upper 2pi. (Not sure if this is correct!)

How did you "get" these limits? What did the instructions tell you to do?

Using the formula, I got the following integral:

. . . . .$\displaystyle \displaystyle \int_0^{6.28}\, \left(3\, +\, 2\sin(x)\right)^2\, \left(-2\sin(x)\right)\, dx$

(with pi before it, of course).

What's "the formula" you're using?

Note: If the integral is set up incorrectly, you may be deriving the correct value for the integral but not for the exercise.

 

[imperatormk]

How did you "get" these limits? What did the instructions tell you to do?

What's "the formula" you're using?

Note: If the integral is set up incorrectly, you may be deriving the correct value for the integral but not for the exercise.

Here is the formula:

. . . . .$\displaystyle V\, =\, \pi\,$$\displaystyle \displaystyle \int_{t_1}^{t_2}\, y^2(t)\, \dot{x}(t)\, dt$

. . . . .

I believe the limits are the problem. I got them not by using a rule or anything, but according to some other exercises I have done.

The integral value based on the limits is 100% correct, but not sure about the limits themselves.

 

[pka]

So I have a circle with the following parametric equations:
. . . . .$\displaystyle \begin{cases}x&=&6\, +\, 2\cos(t)\\y&=&3\, +\, 2\sin(t)\end{cases}$
I have found that the lower limit is 0, and the upper 2pi. (Not sure if this is correct!)
Using the formula, I got the following integral:
. . . . .$\displaystyle \displaystyle \int_0^{6.28}\, \left(3\, +\, 2\sin(x)\right)^2\, \left(-2\sin(x)\right)\, dx$
(with pi before it, of course).
I get the correct value (48pi^2), however, it's negative.

You friend have real problems in understanding what is ask of you.

First, is the volume rotated about the x-axis or about the y-axis ?
In this case it makes a difference as to what method of integration you will use.

Also. If you do not change from parametric to rectangular form (& I would not) then your variable of integration is $\displaystyle \Large t$.

Please replay with what you are trying to do.

 

[imperatormk]

You friend have real problems in understanding what is ask of you.

First, is the volume rotated about the x-axis or about the y-axis ?
In this case it makes a difference as to what method of integration you will use.

Also. If you do not change from parametric to rectangular form (& I would not) then your variable of integration is $\displaystyle \Large t$.

Please replay with what you are trying to do.

Rotated about the X-axis.

It's t I'm actually using, but this online calculator used an x.

 

[pka]

So I have a circle with the following parametric equations:
. . . . .$\displaystyle \begin{cases}x&=&6\, +\, 2\cos(t)\\y&=&3\, +\, 2\sin(t)\end{cases}$

Here is the formula:
. . . . .$\displaystyle V\, =\, \pi\,$$\displaystyle \displaystyle \int_{t_1}^{t_2}\, y^2(t)\, \dot{x}(t)\, dt$
. . . . .

Where in the bloody world did you come up with that formula?
I have done the calculations here. The answer is $\displaystyle 48\pi$ and it is a volume about the x-axis.

First rotate the upper half-circle about the x-axis. Then subtract the rotatation of the lower half-circle about the x-axis.

Here it is Here is the logic $\displaystyle \sin(t)=-\sin(-t)$.

Using that we get: $\displaystyle y^2(t)-y^2(-t)=24\sin(t)$.