# Finding the volume of the sold in the first octant..

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• May 6th 2012, 04:18 PM
mdawg467
Finding the volume of the sold in the first octant..
The question:
Find the volume of the solid in the first octant bound by the graphs of z=1-y^2, y=2x, and x=3.

My attempt:
Attachment 23796
The reason why I changed the limits of integration was because I need x,y,z all to be greater than or equal to 0..however, I am not getting the correct answer of 15/8.
Any suggestions?

Thanks,
Mike.
• May 6th 2012, 07:18 PM
Prove It
Re: Finding the volume of the sold in the first octant..
Your limits are off.

It should be clear that

\displaystyle \begin{align*} 0 \leq x &\leq 3 \\ \\ 0 \leq y &\leq 2x \\ \\ 0 \leq z &\leq 1 - y^2 \end{align*}

So your integral should be

\displaystyle \begin{align*} \int_{0}^{3}{\int_{0}^{2x}{\int_{0}^{1-y^2}{\,dz}\,dy}\,dx} \end{align*}
• May 6th 2012, 07:50 PM
mdawg467
Re: Finding the volume of the sold in the first octant..
Quote:

Originally Posted by Prove It
Your limits are off.

It should be clear that

\displaystyle \begin{align*} 0 \leq x &\leq 3 \\ \\ 0 \leq y &\leq 2x \\ \\ 0 \leq z &\leq 1 - y^2 \end{align*}

So your integral should be

\displaystyle \begin{align*} \int_{0}^{3}{\int_{0}^{2x}{\int_{0}^{1-y^2}{\,dz}\,dy}\,dx} \end{align*}

Thank you for your response.

I originally had my limits set up that way, and my final answer was still off..as it was calculated to be -45.

Maybe there is something wrong with my calculations..
Attachment 23801
any thoughts?
Thanks