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.