Question

Find the volume of the solid cut from the first octant by the surface $z=4-x^2-y$.

Answer 1

The volume of the solid is $\frac{128}{15}$.

Answer 2

The volume $V$ is given by the double integral of the height function $z = 4-x^2-y$ over the region $D$ in the $xy$-plane defined by $x \ge 0$, $y \ge 0$, and $z \ge 0$. The condition $z \ge 0$ implies $4-x^2-y \ge 0$, or $y \le 4-x^2$.

The region of integration $D$ is thus described by $x \ge 0$, $y \ge 0$, and $y \le 4-x^2$. This region is in the first quadrant, bounded by the x-axis, the y-axis, and the parabola $y = 4-x^2$. The x-intercept of this parabola is found by setting $y=0$, which gives $x^2=4$, so $x=2$ (since we are in the first octant).

The limits of integration are $0 \le x \le 2$ and $0 \le y \le 4-x^2$.

The volume integral is: $V = \int_0^2 \int_0^{4-x^2} (4-x^2-y) dy dx$

First, integrate with respect to $y$: $\int_0^{4-x^2} (4-x^2-y) dy = \left[(4-x^2)y - \frac{y^2}{2}\right]_0^{4-x^2}$ $= (4-x^2)(4-x^2) - \frac{(4-x^2)^2}{2} = (4-x^2)^2 - \frac{1}{2}(4-x^2)^2 = \frac{1}{2}(4-x^2)^2$

Now, integrate with respect to $x$: $V = \int_0^2 \frac{1}{2}(4-x^2)^2 dx = \frac{1}{2} \int_0^2 (16 - 8x^2 + x^4) dx$ $V = \frac{1}{2} \left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_0^2$ $V = \frac{1}{2} \left(16(2) - \frac{8(2)^3}{3} + \frac{2^5}{5}\right)$ $V = \frac{1}{2} \left(32 - \frac{64}{3} + \frac{32}{5}\right)$ $V = \frac{1}{2} \left(\frac{32 \times 15 - 64 \times 5 + 32 \times 3}{15}\right)$ $V = \frac{1}{2} \left(\frac{480 - 320 + 96}{15}\right) = \frac{1}{2} \left(\frac{256}{15}\right) = \frac{128}{15}$