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Evaluate the triple integral of $x=y^2$ over the region bounded by $z=x$, $z=0$ and $x=1$ My order of integration was $dx\:dy\:dz$.

I want to calculate the volume of this surface. I solved it for $dz\:dy\:dx$ and it was:

$$V=\int_0^1\int_{-\sqrt{x}}^{\sqrt{x}}\int_{0}^{x}\:dz\:dy\:dx$$

And for $dz\:dx\:dy$ would be this:

$$V=\int_{-1}^{1}\int_{y^2}^{1}\int_{0}^{x}dz\:dx\:dy$$

I tried to solve it and the result is this:

$$V=\int_{0}^{1}\int_{-\sqrt{x}}^{\sqrt{x}}\int_{z}^{1}dx\:dy\:dz + \int_{0}^{1}\int_{-\frac{1}{2}}^{\frac{1}{2}}\int_{y^2}^{1}dx\:dy\:dz$$

But i think its wrong please advice me the best solution .

I wanted to post the shape of this surface in 3-dimensional region but I couldn't because I am new user.

  • 0
    What are the ▒ supposed to be? To get a proper integral sign with limits, enclose \int_0^1 in dollar signs to get $\int_0^1$2012-08-05

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