Negative value for area

Question

I have a problem with the following example:Given the surfaces:$$x^2+y^2+z^2=5, x^2+z^2\ge (y-1)^2$$ find the area bounded with them. First problem appears when I try to sketch this in 3D. For the first one is sphere and that is OK, and the second one should be infinite cone right? (but not sure what part of it left or right, since it is on y axis).

Then for intersection I get $x^2+z^2=1$ and after cilindrical coordinates it is easy to see that $0\le r\le 1$ and $0\le \theta \le 2\pi$. After that, I am calculating the area of cone and sphere seperatly. So for sphere i get $$\sqrt{5}\int\limits_{0}^{2\pi}d\theta \int\limits_{0}^{1}\frac{rdr}{\sqrt{5-r^2}}$$. But this integral has a negative value, so I am obviously mistaking somewhere but not sure where.

BTW i used that $y=\sqrt{5-x^2-z^2}$ from the equation of sphere and than plugged that into formula.

Answer 1

It is easier to do it in spherical coordinates.

Your integral should be

$\displaystyle \int_0^{2\pi} \int_{cos^{-1}({\frac{2}{\sqrt5}})}^{\pi - cos^{-1}({\frac{1}{\sqrt5}})} 5 \sin \phi \, d\phi \, d\theta$

If you see from top ($z-$ axis) in $XY$ plane, this is how the projection looks. This sketch helps to understand the limits of integral.

EDIT: also given you were doing it in cylindrical coordinates, here is the integral in cylindrical coordinates with the correct limits. Please note $-1 \leq y \leq 2$ so you will have to split the integral into two parts, one for surface area above $XZ$ plane and one for below it. For $y = -1, r = 2 \,$ and for $\, y = 2, r = 1$

For above: $\displaystyle \sqrt{5} \int_{0}^{2\pi} \int_{1}^{\sqrt 5}\frac{r}{\sqrt{5-r^2}} \, dr \, d\theta$

For below: $\displaystyle \sqrt{5} \int_{0}^{2\pi} \int_{2}^{\sqrt 5}\frac{r}{\sqrt{5-r^2}} \, dr \, d\theta$

Answer 2

$x^2+y^2+z^2=5$ and $ x^2+z^2= (y-1)^2$ intersect at

$y^2+(y-1)^2=5\to y_1=-1;\;y_2$

The surface is in the picture below and its area is $S=2\pi rh$.

where $r=\sqrt 5$ is the radius of the sphere, $h$ is the height $h=|y_2-y_1|=|2-(-1)|=3$

Therefore $$S=6\sqrt 5\pi$$ which is $67\%$ of the area of the sphere $20\pi$