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Change the order of integration of $$\int_0^6 \int_0^{12-2y}\int_0^{\frac{12-2y-x}{3}} x \, dz \, dx \, dy$$ to $dx\,dy\,dz$

So at first I started with graphing the function, first by looking at the XY plane and then looking at the z function:

enter image description here

So first I need to integrate over $x$ so I "scan" the function for the all x-axis as $y$ and $z$ varies. so the I get $x$ from $0$ to $\frac{12-2y-x}{3}\Rightarrow x=12-2y-3z$ next I need to look at the ZY plane and integrate first over the Y-axis which goes from $0$ to $\frac{12-2y-x}{3}$ but beacuse we are on the ZY plane $x=0$ and we get $y=\frac{12-3z}{3}$ and last $z$ goes from $0$ to $12$

So overall I got:

$$\int_0^{12}\int_0^{\frac{12-3z}{2}}\int_0^{12-2y-3z} x \, dx \, dy \, dz$$

Which is incorrect, where did I get it wrong?

Community
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gbox
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1 Answers1

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From the outer integral, we get $0<y<6$

From the middle integral, we get $x=12-2y$. The intersection on $y$-axis is $6$ (set $x=0$).

From the inner integral, we get $z=\frac{12-2y-x}{3}$. The intersection on z-axis is $4$ (set $x=0$ and $y=0$).

So overall, you should get $\int_0^4\int_0^{6-\frac{3}{2}z}\int_0^{12-2y-3z}x\;dx\,dy\,dz$

Everything in your calculation was correct, except that the intersection on z-axis. You should go back and check that.

Michael Hardy
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Jackson
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