Find the volume of the solid using double integral that is enclosed by the planes $y = x^2$, $z = 0$ and $y+z=1$ in Octant 2

Question

Here's my attempt to the problem.

I first made a graph for the solid. This is what it looks like:

After that, I create the graph for the R (the one that is enclosed by these planes). I determined that the bounds for y is from $x^2$ to 1, while the bounds for x is -1 to 0. As stated in the problem, I only considered the solid created in octant 2.

Finally, I did my integration:

$\int_{-1 }^{0}\int_{x^2 }^{1}(1-y) dy dx $

$\int_{-1 }^{0}(y-\frac{y^2}{2}) |\binom{1}{x^2} dx $

$\int_{-1 }^{0}[(1-\frac{1^2}{2})]-[(x^2)-\frac{(x^2)^2}{2}] dx$

Upon inputting that in my calculator, the answer is 4/15 cubic units.

Is my solution correct?