triple integrals

[calc531]

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

 

[Deleted member 4993]

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

Those are equations of cones - draw a picture of those intersecting cones.

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

 

[calc531]

the book called them paraboloids.
i drew it but my problem is i have trouble visually analyzing things and cannot figure out the boundaries for integration

 

[galactus]

The limits for r can be found by changing the given paraboloid equations to polar. As SK pointed out, the paraboloids intersect. Therefore, change to polar, set them equal and solve for r.

\(\displaystyle x^{2}+y^{2}=r^{2},\;\ 1-x^{2}-y^{2}=1-r^{2}\)

\(\displaystyle r^{2}=1-r^{2}\Rightarrow \;\ r=\pm\frac{1}{\sqrt{2}}\)

So, we have:

\(\displaystyle \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}}\int_{r^{2}}^{1-r^{2}}r(1+cos{\theta})dzdrd{\theta}\)

Don't forget to multiply by r when changing to polar.