Volume of the solid bounded by the cylinder $z = 1-x^2$ and planes, $x=y, y=0,z=0$
$$\int_{y=0}^1\int_{x=y}^1 (1-x^2) \, dx \, dy$$
I don't know if my upper and lower limits are correct. Tell me if this is correct.
Or is it $\int_{y=0}^x\int_{x=1}^y (1-x^2) \, dx \, dy$?
You can project the solid onto the xy - plane. You'll find that the projection onto the xy - plane is a right - angled triangle which is given by $$D = \{ (x, y) \ \epsilon \ R^2 \ | \ 0 \le x \le 1, 0 \le y \le x \}$$ and therefore the solid can be represented as $$ \{ (x, y, z) \ \epsilon \ R^3 \ | \ 0 \le z \le 1 - x^2, (x, y) \ \epsilon \ D \}$$ Then you can evaluate the integral using Fubini's theorem and you'll get $$\int_0^1 \ \left ( \int_0^x \left ( \int_0^{1-x^2} dz \right ) dy \right ) dx = \int_0^1 \left ( \int_0^x (1 - x^2) dy \right ) dx = \iint_D(1-x^2)dxdy$$Since D is a type - II domain as well, you can express D as follows:
$$D = \{ (x, y) \ \epsilon \ R^2 \ | \ 0 \le y \le 1, \ y \le x \le 1 \}$$Thus you'll get $$\iint_D(1-x^2)dxdy = \int_0^1 \left ( \int_y^1(1-x^2)dx \right )dy$$ Hence your expression is correct.