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[SOLVED] Rewriting iterated triple integrals
Rewrite this integral as an equivalent iterated integral in the five other orders.
\int_{0}^{1}\int_{\sqrt{x}}^{1}\int_{0}^{1-y}f(x,y,z) dz dy dx
Ok, so I have the shape drawn out.
The projection on the XY-axis is the region bounded by half the parabola y=\sqrt{x} and the line y=1.
So, 0 \leq x \leq y^{2}, 0 \leq y \leq 1
Or 0 \leq x \leq 1, \sqrt{x} \leq y \leq 1
The projection on the YZ-axis is the triangle bounded by z=0 and the line z=1-y.
So, 0 \leq y \leq 1-z, 0 \leq z \leq 1
Or 0 \leq y \leq 1, 0 \leq z \leq 1-y
The projection on the XZ-axis is another triangle, bounded by z=0 and the line z=1-x
So 0 \leq x \leq 1-z, 0 \leq z \leq 1-x
Or 0 \leq x \leq 1, 0 \leq z \leq 1
When I use those to make my integrals, I get
1) \int_{0}^{1}\int_{0}^{\sqrt{y^{2}}}\int_{0}^{1-x}f(x,y,z) dz dx dy
2) \int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-z}f(x,y,z) dy dz dx
3) \int_{0}^{1}\int_{0}^{1-z}\int_{\sqrt{x}}^{1}f(x,y,z) dy dx dz
4) \int_{0}^{1}\int_{0}^{1-y}\int_{0}^{1-z}f(x,y,z) dx dz dy
5) \int_{0}^{1}\int_{0}^{1-z}\int_{0}^{y^{2}}f(x,y,z) dx dy dz
I substituted 1 for f(x,y,z) and evaluated the integrals. Number 5 was the only one that was equal to the original (1/12).
What did I do wrong? I would really appreciate your help, thanks.
Homework Statement
Rewrite this integral as an equivalent iterated integral in the five other orders.
Homework Equations
\int_{0}^{1}\int_{\sqrt{x}}^{1}\int_{0}^{1-y}f(x,y,z) dz dy dx
The Attempt at a Solution
Ok, so I have the shape drawn out.
The projection on the XY-axis is the region bounded by half the parabola y=\sqrt{x} and the line y=1.
So, 0 \leq x \leq y^{2}, 0 \leq y \leq 1
Or 0 \leq x \leq 1, \sqrt{x} \leq y \leq 1
The projection on the YZ-axis is the triangle bounded by z=0 and the line z=1-y.
So, 0 \leq y \leq 1-z, 0 \leq z \leq 1
Or 0 \leq y \leq 1, 0 \leq z \leq 1-y
The projection on the XZ-axis is another triangle, bounded by z=0 and the line z=1-x
So 0 \leq x \leq 1-z, 0 \leq z \leq 1-x
Or 0 \leq x \leq 1, 0 \leq z \leq 1
When I use those to make my integrals, I get
1) \int_{0}^{1}\int_{0}^{\sqrt{y^{2}}}\int_{0}^{1-x}f(x,y,z) dz dx dy
2) \int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-z}f(x,y,z) dy dz dx
3) \int_{0}^{1}\int_{0}^{1-z}\int_{\sqrt{x}}^{1}f(x,y,z) dy dx dz
4) \int_{0}^{1}\int_{0}^{1-y}\int_{0}^{1-z}f(x,y,z) dx dz dy
5) \int_{0}^{1}\int_{0}^{1-z}\int_{0}^{y^{2}}f(x,y,z) dx dy dz
I substituted 1 for f(x,y,z) and evaluated the integrals. Number 5 was the only one that was equal to the original (1/12).
What did I do wrong? I would really appreciate your help, thanks.