The volume of solid limited by: $\displaystyle 2y^2=x$, $\displaystyle \frac{x}{4}+\frac{y}{2}+\frac{z}{4}=1$, $\displaystyle z=0$ and $\displaystyle y=0$

Answer:

$\displaystyle \frac{17}{5}$

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- Nov 8th 2008, 09:07 AMApprentice123Calculate volume
The volume of solid limited by: $\displaystyle 2y^2=x$, $\displaystyle \frac{x}{4}+\frac{y}{2}+\frac{z}{4}=1$, $\displaystyle z=0$ and $\displaystyle y=0$

Answer:

$\displaystyle \frac{17}{5}$ - Nov 8th 2008, 10:06 AMMathstud28
- Nov 8th 2008, 10:09 AMgalactus
$\displaystyle \int_{0}^{1}\int_{2y^{2}}^{4-2y}(-x-2y+4)dxdy$

$\displaystyle \int_{0}^{1}\int_{2y^{2}}^{4-2y}\int_{0}^{-x-2y+4}dzdxdy$

Now, try doing it by switching the limits of integration. - Nov 8th 2008, 10:11 AMMathstud28
- Nov 8th 2008, 11:32 AMApprentice123
thank you very much

- Nov 10th 2008, 02:37 AMApprentice123
There is a "tutorial" to build the graphic.

- Nov 12th 2008, 02:27 PMApprentice123
- Nov 12th 2008, 02:37 PMMathstud28
- Nov 12th 2008, 02:47 PMApprentice123
- Nov 23rd 2008, 10:47 AMApprentice123
Why not?

$\displaystyle \int_{0}^{2} \int_{2y^2}^{4-2y} \int_{0}^{4-x-2y}dzdxdy$