Electric potential, field and charge density problem check

[DaConfusion]

http://www.physicsforums.com/attachment.php?attachmentid=8198&stc=1&d=1162609042
V – Electric potential

I drew the picture of basically a rod with end points a and –a on the x axis with a point b that sits as well on the x positive axis.

Assuming that is correct, I then am asked to find the electric field using my previous answer on the same point. I did not partially derrive with respect to y or z for the j and k vector components because the original potential equation has no y or z variables which means 0.
http://www.physicsforums.com/attachment.php?attachmentid=8199&stc=1&d=1162609042

Please let me know if the problems are worked out correctly.

My next question is:

Finding the volume charge density in spherical coordinates bounded by:

http://www.physicsforums.com/attachment.php?attachmentid=8200&stc=1&d=1162609042

The formula was given by my teacher as he told us to use that in spherical charge denisty problems. He proved it through a tedious triple integral which I was not able to completly copy down.
The problem I am having is I thought I was correctly setting up the problem but when he was doing a similar problem today on magnetism i noticed his bounds resulted in having each integral with 2-3 parts. Like a to r plus r to 2a and etc. I do not understand.

Please help me, if you guys need more details let me know.

[teclo]

http://www.physicsforums.com/attachment.php?attachmentid=8198&stc=1&d=1162609042
V – Electric potential

I drew the picture of basically a rod with end points a and –a on the x axis with a point b that sits as well on the x positive axis.

Assuming that is correct, I then am asked to find the electric field using my previous answer on the same point. I did not partially derrive with respect to y or z for the j and k vector components because the original potential equation has no y or z variables which means 0.
http://www.physicsforums.com/attachment.php?attachmentid=8199&stc=1&d=1162609042

Please let me know if the problems are worked out correctly.

My next question is:

Finding the volume charge density in spherical coordinates bounded by:

http://www.physicsforums.com/attachment.php?attachmentid=8200&stc=1&d=1162609042

The formula was given by my teacher as he told us to use that in spherical charge denisty problems. He proved it through a tedious triple integral which I was not able to completly copy down.
The problem I am having is I thought I was correctly setting up the problem but when he was doing a similar problem today on magnetism i noticed his bounds resulted in having each integral with 2-3 parts. Like a to r plus r to 2a and etc. I do not understand.

Please help me, if you guys need more details let me know.

none of the links work for me.

[DaConfusion]

http://img.photobucket.com/albums/v412/daconfusion/problem4.jpg

http://img.photobucket.com/albums/v412/daconfusion/problem5.jpg

http://img.photobucket.com/albums/v412/daconfusion/problem6.jpg

[DaConfusion]

all that work is mine so please help, I typed it up on microsoft equation editor 3.0 then pasted into paintbrush and uploaded it as an image.

[teclo]

it's been a year since i've had e&m, but the first part looks ok. i'm kind of confused on the second part, because you're saying a is a variable. i thought a was a constant? i guess it doesn't really matter, because the general formula on the axis would be a---->x

the second one requires different integrals because the charge distribution is different for different regions. so, you'd need an integral for each of those regions to accurately calclulate stuff. i.e. 0->a, a->2a -- each region has a different density. make sense?

i had a rough time in e&m (even if i did get an A), so don't take my word as law.

[DaConfusion]

I see, let me try and get more clarification on the 3rd question. As for the second, a was constant but I have to differentiate with respect to x for the i component. Would I use the source point or field? I considered a to be the source which is technically x so i showed that by differentiating with respect to a.