Trouble with definite integral

[Gameowner]

Homework Statement

$$\int _{0}^{2\,\pi}\!\int _{0}^{1}\! 0.5\,{r}^{ 3.5}\sin \left( \theta \right) {\cos}^{2}+ 0.5\,{r}^{ 2.5}{\sin}^{2}+1/5\,{r}^{5}{\cos}^{2}{ dr}\,{d\theta}$$

Homework Equations

The Attempt at a Solution

So I first integrate the r, which was relatively straight forward, the tricky part was the theta, I had to use the double angle formula? then evaluating, which for the end answer, I got

$$\frac{39}{20}\Pi-\frac{3}{4}$$

which looks kinda wrong to me...

tips and hints appreciated

 

[LCKurtz]

It is bad form to start a new thread instead of continuing the thread that generated the question. Your integrand is wrong. Your integrand should be the dot product of these two vectors:

$$\vec F = \langle r^2\cos\theta\sin\theta,r\sin\theta,r^2\cos^2\theta\rangle$$
$$d\vec S = \langle\frac 1 2 \sqrt r \cos\theta,\frac 1 2 \sqrt r \sin\theta,r\rangle$$

 

[Gameowner]

It is bad form to start a new thread instead of continuing the thread that generated the question. Your integrand is wrong. Your integrand should be the dot product of these two vectors:

$$\vec F = \langle r^2\cos\theta\sin\theta,r\sin\theta,r^2\cos^2\theta\rangle$$
$$d\vec S = \langle\frac 1 2 \sqrt r \cos\theta,\frac 1 2 \sqrt r \sin\theta,r\rangle$$

Hi LCKurtz,

I did try to continue the old thread which generated this question, but no further response was given, and I was not sure whether I'm allowed to double post into my own topic. So I took the chance and posted another thread, I do apologize for that and will remember not to do so next time.

Anyway.

That integrand I posted up to IS the dot product of the 2 vectors you stated above.

But isn't the vector $$d\vec S = \langle\frac 1 2 \sqrt r \cos\theta,\frac 1 2 \sqrt r \sin\theta,r\rangle$$[/QUOTE] suppose to be the normal($$\overline{n}$$)? and Ds is the element of integration?(in this case r dr d(theta))

So the dot product I got is

$$\left[ \begin {array}{c} 0.5\,{r}^{ 2.5} \left( \cos \left( \theta \right) \right) ^{2}\sin \left( \theta \right) \\ \noalign{\medskip} 0.5\,{r}^{ 1.5} \left( \sin \left( \theta \right) \right) ^{2} \\ \noalign{\medskip}{r}^{3} \left( \cos \left( \theta \right) \right) ^{2}\end {array} \right]$$

or more correctly

$$0.5\,{r}^{ 2.5} \left( \cos \left( \theta \right) \right) ^{2}\sin \left( \theta \right) + 0.5\,{r}^{ 1.5} \left( \sin \left( \theta \right) \right) ^{2}+{r}^{3} \left( \cos \left( \theta \right) \right) ^{2}$$

(Note: the reason why I have the coefficients and powers in decimals rather than fractions, is that Maple will evaluate it for me if I were to put it into fractions.)

So after putting in Ds=r dr dtheta, and the limits of integration, I yield the Integrand

$$\int _{0}^{2\,\pi}\!\int _{0}^{1}\! \left( 0.5\,{r}^{ 2.5} \left( \cos \left( \theta \right) \right) ^{2}\sin \left( \theta \right) + 0.5\,{r}^{ 1.5} \left( \sin \left( \theta \right) \right) ^{2}+{r}^{ 3} \left( \cos \left( \theta \right) \right) ^{2} \right) r{dr}\,{d \theta}$$

then I multiplied the r through each term and finally get the Integrand I got up on post 1.

any further tips and hints appreciated, and I apologize again about posting the twice.

Thanks.

 

[LCKurtz]

$$\int _{0}^{2\,\pi}\!\int _{0}^{1}\! \left( 0.5\,{r}^{ 2.5} \left( \cos \left( \theta \right) \right) ^{2}\sin \left( \theta \right) + 0.5\,{r}^{ 1.5} \left( \sin \left( \theta \right) \right) ^{2}+{r}^{ 3} \left( \cos \left( \theta \right) \right) ^{2} \right) r{dr}\,{d \theta}$$

OK, maybe I overlooked something. That last integral is correct. But I thought I had already explained to you how to do the theta integrals: u substitution on the first one and the double angle (or half angle, whichever they are called) formulas for the sin² and cos² ones.

 

[Gameowner]

OK, maybe I overlooked something. That last integral is correct. But I thought I had already explained to you how to do the theta integrals: u substitution on the first one and the double angle (or half angle, whichever they are called) formulas for the sin² and cos² ones.

hi again LCKurtz,

I went back to my old thread and started questions b) and c) again which are located here

https://www.physicsforums.com/showthread.php?p=2672658#post2672658

at which I got these Integrals.

For b), I had to calculate the Integrals for the surface S, which is also S1+S2, therefor I got these as my integrals.

S1

$$\int _{0}^{2\,\pi}\!\int _{0}^{1}\! 0.5\,{r}^{ 3.5} \left( \cos \left( \theta \right) \right) ^{2}\sin \left( \theta \right) + 0.5\, {r}^{ 2.5} \left( \sin \left( \theta \right) \right) ^{2}+{r}^{4} \left( \cos \left( \theta \right) \right) ^{2}{dr}\,{d\theta}$$

S2
$$\int _{0}^{2\,\pi}\!\int _{0}^{1}\!-{r}^{4} \left( \cos \left( \theta \right) \right) ^{2}{dr}\,{d\theta}$$

S1+S2

$$\int _{0}^{2\,\pi}\!\int _{0}^{1}\! 0.5\,{r}^{ 3.5} \left( \cos \left( \theta \right) \right) ^{2}\sin \left( \theta \right) + 0.5\, {r}^{ 2.5} \left( \sin \left( \theta \right) \right) ^{2}{dr}\,{d \theta}$$

And for the volume Integral c), I got

$$\int _{0}^{2\,\pi}\!\int _{0}^{1}\! 0.5\,{r}^{ 3.5} \left( \cos \left( \theta \right) \right) ^{2}\sin \left( \theta \right) + 0.5\, {r}^{ 2.5} \left( \sin \left( \theta \right) \right) ^{2}{dr}\,{d \theta}$$


The question states that the volume integral should be the same as the integral of S1+S2 when I have evaluated it, but when I do, it does not give me the same answer?

any help, hints would be much appreciated.

 

[LCKurtz]

OK, maybe I overlooked something. That last integral is correct. But I thought I had already explained to you how to do the theta integrals: u substitution on the first one and the double angle (or half angle, whichever they are called) formulas for the sin² and cos² ones.

The equation I was referring to was:
$$\int _{0}^{2\,\pi}\!\int _{0}^{1}\! \left( 0.5\,{r}^{ 2.5} \left( \cos \left( \theta \right) \right) ^{2}\sin \left( \theta \right) + 0.5\,{r}^{ 1.5} \left( \sin \left( \theta \right) \right) ^{2}+{r}^{ 3} \left( \cos \left( \theta \right) \right) ^{2} \right) r{dr}\,{d \theta}$$

When I told you it was correct I overlooked that r outside the parentheses. It should not be there. You don't need that extra r when you use the cross product to calculate dS.

hi again LCKurtz,

I went back to my old thread and started questions b) and c) again which are located here

https://www.physicsforums.com/showthread.php?p=2672658#post2672658

at which I got these Integrals.

For b), I had to calculate the Integrals for the surface S, which is also S1+S2, therefor I got these as my integrals.

S1

$$\int _{0}^{2\,\pi}\!\int _{0}^{1}\! 0.5\,{r}^{ 3.5} \left( \cos \left( \theta \right) \right) ^{2}\sin \left( \theta \right) + 0.5\, {r}^{ 2.5} \left( \sin \left( \theta \right) \right) ^{2}+{r}^{4} \left( \cos \left( \theta \right) \right) ^{2}{dr}\,{d\theta}$$

S2
$$\int _{0}^{2\,\pi}\!\int _{0}^{1}\!-{r}^{4} \left( \cos \left( \theta \right) \right) ^{2}{dr}\,{d\theta}$$

Both of the above integrals have that extra r in them. Remove the extra r and they should both give you the correct answer.

S1+S2

<snip> Calculate them separately.

And for the volume Integral c), I got

$$\int _{0}^{2\,\pi}\!\int _{0}^{1}\! 0.5\,{r}^{ 3.5} \left( \cos \left( \theta \right) \right) ^{2}\sin \left( \theta \right) + 0.5\, {r}^{ 2.5} \left( \sin \left( \theta \right) \right) ^{2}{dr}\,{d \theta}$$

The volume integral shouldn't be nearly that complicated and it should be a triple integral, not a double integral. Set up the triple integral of [itex] \nabla \cdot F [/tex] for the enclosed volume in cylindrical coordinates and post it.

 

[Gameowner]

The equation I was referring to was:
$$\int _{0}^{2\,\pi}\!\int _{0}^{1}\! \left( 0.5\,{r}^{ 2.5} \left( \cos \left( \theta \right) \right) ^{2}\sin \left( \theta \right) + 0.5\,{r}^{ 1.5} \left( \sin \left( \theta \right) \right) ^{2}+{r}^{ 3} \left( \cos \left( \theta \right) \right) ^{2} \right) r{dr}\,{d \theta}$$

When I told you it was correct I overlooked that r outside the parentheses. It should not be there. You don't need that extra r when you use the cross product to calculate dS.



Both of the above integrals have that extra r in them. Remove the extra r and they should both give you the correct answer.



<snip> Calculate them separately.



The volume integral shouldn't be nearly that complicated and it should be a triple integral, not a double integral. Set up the triple integral of [itex] \nabla \cdot F [/tex] for the enclosed volume in cylindrical coordinates and post it.

The extra r you're speaking of came from the element of volume dS which is r dr dtheta? then I multiplied through the r into the integral? or was I not meant to do that?

lastly, I have no idea why the integral came out on here the way it has, but this was my triple integral:

$$\int _{0}^{2\,\pi}\!\int _{0}^{1}\!\int _{1-\sqrt {r}}^{0}\!{r}^{3} \cos \left( \theta \right) \left( \sin \left( \theta \right) \right) ^{2}+r\sin \left( \theta \right) {dz}r{dr}\,{d\theta}$$

 

[LCKurtz]

The extra r you're speaking of came from the element of volume dS which is r dr dtheta? then I multiplied through the r into the integral? or was I not meant to do that?

In general when you have a surface defined parametrically as

$$\vec R(u,v) = \langle x(u,v), y(u,v), z(u,v)\rangle$$

the element of surface are is defined by

$$dS = |\vec R_u \times \vec R_v|\, dudv$$

with no extra u or v. In the case of polar coordinates we get used to writing $rdrd\theta$ as a shortcut. But when you have a surface parameterized as $R(r,\theta)$ and calculate

$$dS = |\vec R_r \times \vec R_\theta|$$

the r gets built in. Try it with polar coordinates in the plane, which you would parameterize as

$$\vec R(r,\theta) = \langle r\cos\theta, r\sin\theta, 0\rangle$$

Calculate dS from that and you will see the r is built in. That is in fact where the r we love so well in the formula $rdrd\theta$ comes from.

lastly, I have no idea why the integral came out on here the way it has, but this was my triple integral:

$$\int _{0}^{2\,\pi}\!\int _{0}^{1}\!\int _{1-\sqrt {r}}^{0}\!{r}^{3} \cos \left( \theta \right) \left( \sin \left( \theta \right) \right) ^{2}+r\sin \left( \theta \right) {dz}r{dr}\,{d\theta}$$

If I recall correctly, the vector field you started with was F = <xy, y, x²>. When you take its divergence and parameterize it you don't get anything nearly that complicated. And are your limits in the positive direction?

 

[Gameowner]

In general when you have a surface defined parametrically as

$$\vec R(u,v) = \langle x(u,v), y(u,v), z(u,v)\rangle$$

the element of surface are is defined by

$$dS = |\vec R_u \times \vec R_v|\, dudv$$

with no extra u or v. In the case of polar coordinates we get used to writing $rdrd\theta$ as a shortcut. But when you have a surface parameterized as $R(r,\theta)$ and calculate

$$dS = |\vec R_r \times \vec R_\theta|$$

the r gets built in. Try it with polar coordinates in the plane, which you would parameterize as

$$\vec R(r,\theta) = \langle r\cos\theta, r\sin\theta, 0\rangle$$

Calculate dS from that and you will see the r is built in. That is in fact where the r we love so well in the formula $rdrd\theta$ comes from.
If I recall correctly, the vector field you started with was F = <xy, y, x²>. When you take its divergence and parameterize it you don't get anything nearly that complicated. And are your limits in the positive direction?

Oh! Thank you very much for clearing that up for me, we never got taught when there is no r in dV, we never really get taught that stuff, basically just the way to approach questions, which I'm still not very good at.

Anyway, in regards to the triple integral, I calculated it as it follows.

$$\nabla$$ = $$\left[\frac{d}{dx}\frac{d}{dy}\frac{d}{dz}\right]$$

= $$\left[\frac{d(xy)}{dx}\frac{d(y)}{dy}\frac{d(x^2)}{dz}\right]$$

=$$\left[\,y,1,0\right]$$

then my F is

$$\left[ \begin {array}{c} xy\\ \noalign{\medskip}y \\ \noalign{\medskip}{x}^{2}\end {array} \right]$$

taking the dot product of the 2, gives me

Triple Integral (x*y^2+y) dV

and substituted the cylindrical coordinates x=rcos(theta), y=rsin(theta) into the triple integral, and thus yielding the triple integral I posted above.

was I suppose to make the substitution BEFORE the dot product?

Thanks for your reply.

 

[LCKurtz]

You are supposed to calculate $\nabla \cdot F$. This is a dot product which is a scalar, not a vector.

 

[Gameowner]

You are supposed to calculate $\nabla \cdot F$. This is a dot product which is a scalar, not a vector.

Oh right...so I would just have 1+y then wouldn't I?

 

[LCKurtz]

Oh right...so I would just have 1+y then wouldn't I?

Bingo. Now remember the correct formula for dV in cylindrical coordinates. Unlike the surface differential, you aren't calculating it directly from a parameterization which builds in the r.

At this point hopefully you can get the correct numbers for both sides of the divergence theorem.

 

[Gameowner]

Bingo. Now remember the correct formula for dV in cylindrical coordinates. Unlike the surface differential, you aren't calculating it directly from a parameterization which builds in the r.

At this point hopefully you can get the correct numbers for both sides of the divergence theorem.

Great! I now have S1+S2=Volume Integral,

However, for S1+S2, I got a positive answer and for the volume integral I got a negative answer. I assume the limits for my volume integral are in the wrong direct? if so, how will I know which direct they're suppose to be and how do I fix this?

Thanks.

 

[LCKurtz]

Make sure you integrated in the positive direction for all three variables in the volume integral.