# Evaluate the triple integral.

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.

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What are the ▒ supposed to be? To get a proper integral sign with limits, enclose \int_0^1 in dollar signs to get $\int_0^1$ –  Ross Millikan Aug 5 '12 at 21:18

Integrating in three dimensions will give you a volume, not the area of a surface. Your region is not well defined in the first line-it is a triangle in the $xz$ plane but there is no restriction in the $y$ direction. If you want the region to be bounded by $x=y^2$ then your integral is correct, $V=\int_0^1\int_{-\sqrt{x}}^{\sqrt{x}}\int_{0}^{x}\:dz\:dy\:dx=\int_0^1\int_{-\sqrt{x}}^{\sqrt{x}}x\:dy\:dx=\int_0^12x\sqrt{x}\:dx=\frac 45 x^{\frac 52}|_0^1=\frac 45$. This is the triple integral of $1$ over that volume.

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yes it would be 4/5 for dzdydx & dzdxdy but for the dxdydz with that formula that i wrote its not the same so its wrong i don't know how to fix it –  shahin Aug 6 '12 at 0:37
for $dx \:dy \:dz$, the lower $x$ limit has to be the maximum of $y^2$ and $z$, you don't add the two integrals together. In your last line, you can't have $x$ in the limits of the $y$ integral as you have already integrated over $x$. In the second integral on that line, I don't know where the $y$ limits came from-we don't have any $\frac 12$'s around. –  Ross Millikan Aug 6 '12 at 0:50
yes i think it would be $$\int_0^1\int_{-1}^{1 }\int_1^{y^2}dxdydz$$ –  shahin Aug 6 '12 at 1:02
but its still wrong –  shahin Aug 6 '12 at 1:28