# Math Help - Finding the volume of the sold in the first octant..

1. ## 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:

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.

2. ## 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*}

3. ## Re: Finding the volume of the sold in the first octant..

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..

any thoughts?
Thanks