We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the \(\displaystyle xy\) plane, this means we are above \(\displaystyle z=0\).

\(\displaystyle 0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1\)

The region \(\displaystyle D\) is between two circles \(\displaystyle x^2+y^2=1\), and \(\displaystyle x^2+y^2=9\).

This means that 

\(\displaystyle 0 \leq \theta \leq 2\pi\)

\(\displaystyle 1\leq r \leq 3\)

\(\displaystyle \int \int \int_D y \ dV=\int_{0}^{2\pi} \int_{1}^{3} \int_{0}^{r\cos(\theta)+1} r\sin(\theta) r\ dz \ dr \ d\theta\)

\(\displaystyle =\int_{0}^{2\pi} \int_{1}^{3} \int_{0}^{r\cos(\theta)+1} r^2\sin(\theta) \ dz \ dr \ d\theta\)

\(\displaystyle =\int_{0}^{2\pi} \int_{1}^{3} zr^2\sin(\theta) \Big|_{0}^{r\cos(\theta)+1} \ dr \ d\theta\)

\(\displaystyle =\int_{0}^{2\pi} \int_{1}^{3} (r\cos(\theta)+1)r^2\sin(\theta)-0 \ dr \ d\theta\)

\(\displaystyle =\int_{0}^{2\pi} \int_{1}^{3} r^3\sin(\theta)\cos(\theta)+r^2\sin(\theta) \ dr \ d\theta\)

\(\displaystyle =\int_{0}^{2\pi} \frac{1}{4}r^4\sin(\theta)\cos(\theta)+\frac{1}{3}r^3\sin(\theta) \Big|_{1}^{3}\ d\theta\)

\(\displaystyle =\int_{0}^{2\pi} (\frac{1}{4}(3)^4\sin(\theta)\cos(\theta)+\frac{1}{3}(3)^3\sin(\theta))-(\frac{1}{4}(1)^4\sin(\theta)\cos(\theta)+\frac{1}{3}(1)^3\sin(\theta)) \ d\theta\)

\(\displaystyle =\int_{0}^{2\pi} (\frac{81}{4}\sin(\theta)\cos(\theta)+9\sin(\theta))-(\frac{1}{4}\sin(\theta)\cos(\theta)+\frac{1}{3}\sin(\theta)) \ d\theta\)

\(\displaystyle =\int_{0}^{2\pi} 20\sin(\theta)\cos(\theta)+\frac{26}{3}\sin(\theta) \ d\theta\)

\(\displaystyle = -\frac{1}{2}\cdot20\cos^2(\theta)-\frac{26}{3}\cos(\theta)\Big|_{0}^{2\pi}\)

\(\displaystyle = -\frac{1}{2}\cdot20\cos^2(2\pi)-\frac{26}{3}\cos(2\pi)-(-\frac{1}{2}\cdot20\cos^2(0)-\frac{26}{3}\cos(0))\)

\(\displaystyle = -\frac{1}{2}\cdot20(1)-\frac{26}{3}(1)-(-\frac{1}{2}\cdot20(1)-\frac{26}{3}(1))\)

\(\displaystyle = -10-\frac{26}{3}-(-10-\frac{26}{3})=0\)

Remember the general equation of a line in vector form:

\(\displaystyle \vec{r}=r_0+t\vec{v}=\left \langle x_0,y_0,z_0\right \rangle+t\left \langle a,b,c\right \rangle\), where \(\displaystyle \left \langle x_0,y_0,z_0\right \rangle\) is the starting point, and \(\displaystyle \left \langle a,b,c\right \rangle\) is the difference between the start and ending points.

Lets apply this to our problem.

\(\displaystyle \vec{r}=\left \langle 2, 10, -6\right \rangle+t\left \langle 5, 6, -20\right \rangle\)

Distribute the \(\displaystyle t\)

\(\displaystyle \vec{r}=\left \langle 2, 10, -6\right \rangle+\left \langle 5t, 6t, -20t\right \rangle\)

Now we simply do vector addition to get

\(\displaystyle \vec{r}=\left \langle 2+5t, 10+6t, -6-20t\right \rangle\)