Ok. So you have the triple integral:
$$\begin{align}
\int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\int_1^{5-y} \;dz\;dx\;dy
&= \int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}4-y\;dx\;dy \\
&=\int_{-3}^34x-xy\Bigg|_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\;dy \\
&=\int_{-3}^38\sqrt{9-y^2}-2y\sqrt{9-y^2}\;dy \\
&= 8\int_{-3}^33\sqrt{1-\left(\frac{y}{3}\right)^2}\;dy-2\int_{-3}^3y\sqrt{9-y^2}\;dy
\end{align}$$
Now, I'm going to break this up. For the left-hand integral, we must use trig-substitution. Let $\cos(t) = \frac{y}{3}$. This implies that $dy = -3\sin(t)\;dt$. The limits of integration change as well, to $t=\arccos\left(\frac{-3}{3}\right) = \pi$ to $t = \arccos\left(\frac{3}{3}\right) = 0$.
So, the integral becomes:
$$\begin{align}
24\int_\pi^0\sqrt{1-\cos^2(t)}(-3\sin t)\;dt &= -72\int_\pi^0\sin^2(t)\;dt\\
&=72\int_0^\pi\frac{1}{2}-\frac{\cos(2t)}{2}\;dt\\
&=36\int_0^\pi1-\cos(2t)\;dt\\
&=36\left(t - \frac{\sin(2t)}{2}\right)\Bigg|_0^\pi\\
&=\boxed{36\pi}
\end{align}$$
Now, for the left-hand integral, we apply $u$-substitution. If we set $u = 9-y^2$, then $du = -2y\;dy$. The limits are transformed to $u = 9-(-3)^2 = 0$ to $u = 9-(3)^2 = 0$
So, the integral becomes:
$$\begin{align}
-2\int_{-3}^3y\sqrt{9-y^2}\;dy &= \int_0^0\sqrt{u}\;du\\
&= \boxed{0}
\end{align}$$
Well, that wasn't exciting. :)
So, putting it all together, we end up with:
$$V = 36\pi + 0 = \boxed{36\pi}$$
The solution appears to be correct. Personally, I used different construction of the integral, which is
$\int_0^2dy\int_0^{2-y}dz\int_0^{4-y^2}dx = \frac{20}{3}$.
Hope this helps.
Best Answer
Here is an alternative computation using a single variable integral that confirms your result. The following figure represents the given pyramid.
The equations of the lines situated on the planes $y=0$ and $z=0$ are:
$$y=0,\qquad\frac{x}{a}+\frac{z}{c}=1\Leftrightarrow z=\left( 1-\frac{x}{a}\right) c,$$
$$z=0,\qquad\frac{x}{a}+\frac{y}{c}=1\Leftrightarrow y=\left( 1-\frac{x}{a}\right) b.$$
The intersection of the pyramid with the plane perpendicular to the $x$-axis in $x$ is a right triangle with catheti $\left( 1-\frac{x}{a}\right) c$ and $\left( 1-\frac{x}{a}\right) b$, whose area $A(x)$ is given by
$$A(x)=\frac{1}{2}\left( 1-\frac{x}{a}\right) b\left( 1-\frac{x}{a}\right) c=\frac{bc}{2}\left( 1-\frac{x}{a}\right) ^{2}.$$
Hence the volume is given by the integration of the area $A(x)$ from $x=0$ to $x=a$
$$\begin{eqnarray*} V &=&\int_{0}^{a}A(x)dx \\ &=&\frac{bc}{2}\int_{0}^{a}\left( 1-\frac{x}{a}\right) ^{2}dx \\ &=&\frac{bc}{2}\left( a-\frac{2}{a}\frac{a^{2}}{2}+\frac{1}{a^{2}}\frac{a^{3} }{3}\right) \\ &=&\frac{abc}{6}. \end{eqnarray*}$$