Find the volume of the solid bounded by the cylinder x² +

Find the volume of the solid bounded by the cylinder $x^2 + y^2 = a^2$ and the planes $z = 0$ and $y + z = 0$.

We integrate the wedge-shaped height $-y$ over the lower half of the disk, first in Cartesian and then in polar coordinates as a check.

Step 1: Determine the base region

The lower bound is $z=0$; the upper bound comes from $y+z=0 \implies z=-y$. Since we need $z\ge0$, this forces $-y\ge0$, i.e. $y\le0$ — the solid sits over the lower half of the disk $x^2+y^2\le a^2$.

Step 2: Set up the volume integral

The height at $(x,y)$ is $h(x,y)=-y$, so:

\[V = \iint_D(-y)\,dA, \qquad D=\{x^2+y^2\le a^2,\ y\le0\}\]

Step 3: Evaluate in Cartesian coordinates

\[V = \int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{0}(-y)\,dy\,dx\]

Inner integral: $\left[-\dfrac{y^2}{2}\right]_{-\sqrt{a^2-x^2}}^{0} = \dfrac{a^2-x^2}{2}$.

\[V = \frac12\int_{-a}^{a}(a^2-x^2)\,dx = \frac12\cdot\frac{4a^3}{3} = \frac{2a^3}{3}\]

\[\boxed{V = \frac{2a^3}{3} \text{ cubic units}}\]

Step 4: Cross-check in polar coordinates

Here $y\le0$ corresponds to $\theta\in[\pi,2\pi]$, and $h=-r\sin\theta$:

\[V = \int_{\pi}^{2\pi}\int_0^{a}(-r\sin\theta)r\,dr\,d\theta = -\frac{a^3}{3}\int_{\pi}^{2\pi}\sin\theta\,d\theta = -\frac{a^3}{3}[-\cos\theta]_{\pi}^{2\pi} = -\frac{a^3}{3}(-2) = \frac{2a^3}{3} \quad\checkmark\]

Summary

\[\boxed{V = \frac{2a^3}{3} \text{ cubic units}}\]

Key insight: The tilted plane $y+z=0$ slices only the lower half of the cylinder (since it would require negative height over the upper half), creating a wedge whose volume is exactly $\frac{2a^3}{3}$ — independent of $\pi$, since the varying wedge height cancels the circular cross-section's usual $\pi$ factor.