The problem is faulty: it gives you only one boundary of the region and leaves you to guess about the others. There are two possible regions. One, which I’ll call $R_1$, is bounded by $y=\cos x$, the segment of the $y$-axis between $y=0$ and $y=1$, and the segment of the $x$-axis between $x=0$ and $x=\frac{\pi}2$; it’s the region whose area you would be finding if you were asked to find the area under $y=\cos x$ between $x=0$ and $x=\frac{\pi}2$. The other, which I’ll call $R_2$, is bounded by $y=\cos x$, the segment of the line $x=\frac{\pi}2$ between $y=0$ and $y=1$, and the segment of the line $y=1$ between $x=0$ and $x=\frac{\pi}2$. My best guess is that you’re intended to use $R_1$, so that’s the one that I’ll use.
If you revolve $R_2$ about the line $y=1$, you get a roughly cone-shaped solid; its peak is to the left, at the point $\langle 0,1\rangle$, and its base is to the right. Its height, measured horizontally from peak to base, is $\frac{\pi}2$; its base has radius $1$. If you revolve $R_2$ about the line $y=1$, the resulting solid bears no resemblance to a cone.
If you chop $R_1$ into vertical slices, each slice produces an annulus. The outer radius of the annulus is the distance from the $x$-axis to the axis of revolution, which is the line $y=1$, so the outer radius is $1$. The inner radius is the distance from the top edge of $R_1$, the line $y=\cos x$, to the axis of revolution, so it’s $1-\cos x$. The area of the annulus is therefore
$$\pi\left(1^2-(1-\cos x)^2\right)=\pi\left(2\cos x-\cos^2x\right)\;,$$
and it contributes $$dV=\pi\left(2\cos x-\cos^2x\right)dx$$ to the volume. There is one such slice for each $x$ from $0$ to $\frac{\pi}2$, so the volume is
$$V=\pi\int_0^{\pi/2}\left(2\cos x-\cos^2x\right)dx\;.$$
Using cylindrical shells for this problem is a bad idea, but it can be done. Since you’re revolving the region about a horizontal axis, you must slice it up horizontally $-$ parallel to the axis of revolution $-$ in order to get shells. This means that you’ll have a shell for each value of $y$ from $0$ to $1$, and the infinitesimal thickness of each shell will be $dy$, not $dx$. (Infinitesimal here is informal, but it’s a perfectly good way to think about the problem in order to get it set up right.) The radius of the shell at a particular value of $y$ is the distance from $y$ to the axis of revolution at $y=1$, which is $1-y$. The height of the shell is the horizontal distance from the $y$-axis to the curve $y=\cos x$, which is $\cos^{-1}y$. The shell therefore contributes
$$dV=2\pi(1-y)\cos^{-1}y\,dy$$
to the volume of the solid of revolution, and the whole volume is
$$V=2\pi\int_0^1(1-y)\cos^{-1}y\,dy\;.$$
Best Answer
The reason you need two integrals is because over the range $0$ to $4$ in $y$, the height of the shell is $2\sqrt y$ because it extends all the way across the $y$ axis. Over the range $4$ to $9$ in $y$, the height is $\sqrt y -(y-6)$. That changes the integrand, so split it into two integrals.
A sketch is below. The horizontal lines are what are revolved to make the shells. Note that the lower one extends from one side of the parabola to the other. The upper one extends from the line to the parabola. That is why the integral changes