# Homework Help: Trouble with definite integral

1. Apr 18, 2010

### Gameowner

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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

2. Apr 18, 2010

### LCKurtz

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

3. Apr 18, 2010

### Gameowner

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.

4. Apr 19, 2010

### 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.

5. Apr 19, 2010

### Gameowner

hi again LCKurtz,

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

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.

6. Apr 20, 2010

### LCKurtz

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 $\nabla \cdot F [/tex] for the enclosed volume in cylindrical coordinates and post it. 7. Apr 20, 2010 ### Gameowner 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: $$\int _{0}^{2\,\pi}\!\int _{0}^{1}\!\int _{1-\sqrt {r}}^{0}\!{r}^{3} \cos \left( \theta \right) \left( \sin \left( \theta \right) \right) ^{2}+r\sin \left( \theta \right) {dz}r{dr}\,{d\theta}$$ 8. Apr 20, 2010 ### LCKurtz 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 [itex]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?

9. Apr 20, 2010

### Gameowner

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?

10. Apr 20, 2010

### LCKurtz

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

11. Apr 21, 2010

### Gameowner

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

12. Apr 21, 2010

### LCKurtz

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.

13. Apr 21, 2010

### Gameowner

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.

14. Apr 21, 2010

### LCKurtz

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