Chapter 15 - Practice Exercises - Page 909: 6

$5$

Here, we have $x=1+9t \implies dx=9 dt, y= 1+2t \implies dy=2 dt$ and $z=1+2t \implies dz=2dt$ Plug the above values in the given integral. $\int_{0}^{1} 9dt-\sqrt{\dfrac{1+2t}{1+2t}}(2 dt) -\sqrt{\dfrac{1+2t}{1+2t}}(2 dt)$ Thus, we have $\int_{0}^{1}9 dt-2 dt -2dt=5\int_0^1 dt=5$