# I do not understand or get these problems?

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• Oct 15th 2010, 01:22 PM
elpermic
I do not understand or get these problems?
http://i56.tinypic.com/fjdqw2.png

Yeah.. Homework help. I seriously do not understand it even though I am following exact procedures and made sure I wasn't wrong, but I'm still wrong? I'm on my last submission for my online homework in order to get points and I would like to get it right.. But I don't understand it?
• Oct 15th 2010, 01:32 PM
mr fantastic
Quote:

Originally Posted by elpermic
http://i56.tinypic.com/fjdqw2.png

Yeah.. Homework help. I seriously do not understand it even though I am following exact procedures and made sure I wasn't wrong, but I'm still wrong? I'm on my last submission for my online homework in order to get points and I would like to get it right.. But I don't understand it?

Please show all your working so that you can be pointed in the right direction.
• Oct 15th 2010, 01:37 PM
elpermic
well for the 1st one..

i take the integral of π r^2 from [0,2] and I get 8.378 as my answer(im sure this is right.. but im wrong)

as for the 2nd one

i take the integral of 4(3-3x^2) from [-1,1] and after calculation i get 24. still wrong though
• Oct 15th 2010, 01:42 PM
mr fantastic
Quote:

Originally Posted by elpermic
well for the 1st one..

i take the integral of π r^2 from [0,2] and I get 8.378 as my answer(im sure this is right.. but im wrong)

as for the 2nd one

i take the integral of 4(3-3x^2) from [-1,1] and after calculation i get 24. still wrong though

I will point you in the right direction for #1:

$\displaystyle \displaystyle V = \pi \int_3^5 x^2 \, dy$. An exact answer is probably required.
• Oct 15th 2010, 01:50 PM
elpermic
Sorry. I forgot to say I tried that integral as well and I got 98π/3.. But my answer was still wrong.
• Oct 15th 2010, 01:53 PM
mr fantastic
Quote:

Originally Posted by elpermic
Sorry. I forgot to say I tried that integral as well and I got 98π/3.. But my answer was still wrong.

In my first reply I bold faced the word "all" for a very good reason. Show all your work, every step.
• Oct 15th 2010, 01:56 PM
elpermic
those were the only 2 integrals I tried. Both of them are wrong and Im still confused because I'm sure I did everything right.
• Oct 15th 2010, 02:08 PM
elpermic
I finally got the answer to the first problem! :) But I still can't do the second one.
• Oct 16th 2010, 04:12 PM
mr fantastic
Quote:

Originally Posted by elpermic
I finally got the answer to the first problem! :) But I still can't do the second one.

I will point you in the right direction: $\displaystyle \displaystyle V = \int_0^3 (2x)^2 \, dy$.

Your job is to understand where this has come from and what to do with it.