[SOLVED] Volume (triple integral)

**Spec** · August 9th 2008, 04:54 PM

Find the volume of $D= \left\{(x,y,z)\in R^3: z \geq y^2;\ y\geq x^2;\ x \geq z^2 \right\}$

I'm not quite sure where to start. Do I need to split D up into three different volumes? The body is symmetrical, so I would only need to calculate one of the three volumes in that case.

One of the three volumes is: $x^2 \leq y \leq \sqrt{z}$ , but that doesn't really help me without any boundaries for y.

**NonCommAlg** · August 9th 2008, 06:27 PM

Originally Posted by Spec

Find the volume of $D= \left\{(x,y,z)\in R^3: z \geq y^2;\ y\geq x^2;\ x \geq z^2 \right\}$

I'm not quite sure where to start. Do I need to split D up into three different volumes? The body is symmetrical, so I would only need to calculate one of the three volumes in that case.

One of the three volumes is: $x^2 \leq y \leq \sqrt{z}$ , but that doesn't really help me without any boundaries for y.

put: $\frac{x^2}{y}=u, \ \frac{y^2}{z}=v, \ \frac{z^2}{x}=w.$ then your region will be transfered to the unit cube $0 \leq u \leq 1, \ 0 \leq v \leq 1, \ 0 \leq w \leq 1.$

a simple computation shows that $\frac{1}{J}=\frac{\partial(u,v,w)}{\partial(x,y,z) }=7.$ thus: $\text{the volume of D} =\frac{1}{7} \int_0^1 \int_0^1 \int_0^1 du \ dv \ dw = \frac{1}{7}. \ \ \ \square$

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