1. ## Calculate 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 $\displaystyle y=0$

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

2. Originally Posted by Apprentice123
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 $\displaystyle y=0$

$\displaystyle \frac{17}{5}$
Maybe I am making a computational error but I keep getting $\displaystyle \frac{81}{5}$.

3. $\displaystyle \int_{0}^{1}\int_{2y^{2}}^{4-2y}(-x-2y+4)dxdy$

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

Now, try doing it by switching the limits of integration.

4. Originally Posted by galactus
$\displaystyle \int_{0}^{1}\int_{2y^{2}}^{4-2y}(-x-2y+4)dxdy$

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

Now, try doing it by switching the limits of integration.
Ahh thank you very much Galctus! I had $\displaystyle \int_{-2}^{1}\cdots$ for some reason haha

EDIT: I see what I did, I solved the inequality wrong by mistake. It's good now.

5. thank you very much

6. There is a "tutorial" to build the graphic.

7. Originally Posted by Mathstud28
Ahh thank you very much Galctus! I had $\displaystyle \int_{-2}^{1}\cdots$ for some reason haha

EDIT: I see what I did, I solved the inequality wrong by mistake. It's good now.

$\displaystyle 2y^2=x$ A parable

$\displaystyle \frac{x}{4}+\frac{y}{2}+\frac{z}{4}=1$ What is this graphic?

8. Originally Posted by Apprentice123
$\displaystyle 2y^2=x$ A parable

$\displaystyle \frac{x}{4}+\frac{y}{2}+\frac{z}{4}=1$ What is this graphic?
If I am understanding you correctly, a plane.

9. Originally Posted by Mathstud28
If I am understanding you correctly, a plane.

Thank you. Why the integral range from 0 to 1?

10. Why not?

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