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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.

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

Your limits are off.

It should be clear that

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

So your integral should be

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

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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

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

So your integral should be

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

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