# Thread: triple integral

1. ## triple integral

evaluate the triple integral by switching to cylindrical coordinates.
the triple integral of 1+ (x)/(sqrt(x^2+y^2))
over S which is the solid bounded by the paraboloids z=x^2+y^2 and z=1-x^2-y^2

i changed the equation to cylindrical coordinates and simplified to get 1+cos(theta)
my main problem is finding the boundaries to integrate from.
does anyone know how i can find the boundaries for r, theta, and z?

the final answer in the back of the book is pi/4 if that helps.
thank you so much! i just can't figure out how to find the boundaries when it's not in rectangular coordinates..

2. First, we find out where the right-side-up paraboloid $\displaystyle z=x^2+y^2$ and the upside-down paraboloid $\displaystyle z=1-x^2-y^2$ intersect. In doing this, we switch to cylindrical coordinates:

\displaystyle \begin{aligned} x^2+y^2&=1-x^2-y^2\\ r^2&=1-r^2\\ 2r^2&=1\\ r^2&=\frac{1}{2}\\ r&=\frac{1}{\sqrt{2}}. \end{aligned}

The paraboloids therefore intersect in a circle of radius $\displaystyle \frac{1}{\sqrt{2}}$. As the region extends from $\displaystyle z=0$ to $\displaystyle z=1$, we may set up our integral as follows, remembering to include the extra factor $\displaystyle r$:

$\displaystyle \int_0^{2\pi}\int_0^{\frac{1}{\sqrt{2}}}\int_0^1 (1+\cos \theta)r\,dz\,dr\,d\theta.$

Incidentally, I did not obtain $\displaystyle \frac{\pi}{4}$ when I worked out this problem.