Let $$M=\{(x,y,z)\in \mathbb R^3 \mid z=1-x^2-y^2, z>0\}$$ be a two-dimensional submanifold.
Now I need to integrate $$f(x,y,z)=\sqrt{\frac{1}{4}+x^2+y^2}$$ on $M$.
I have chosen $$\phi(\alpha,\beta):=(\cos\alpha\cos\beta,\sin\alpha\cos\beta,\sin^2\beta)$$ as a parametrization of $M$. Calculating the Gramian determinant I obtained $$\det((\phi'(\alpha,\beta))^T \phi'(\alpha,\beta))=(\sin^2\beta+4\cos^2\beta\sin^2\beta)(1+\cos^2\beta).$$ Applying it to $$\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}f(\phi(\alpha,\beta))\sqrt{\det((\phi'(\alpha,\beta))^T \phi'(\alpha,\beta))}\,d\beta\,d\alpha$$ I get weird results, so I'm wondering if I made a mistake until this point? Thanks in advance.
Your Gramian determinant is incorrect. To retain your definition of volume element $\phi'$ must be defined as the transpose of Jacobian or else the volume element should be $\sqrt{\det(\phi'\phi'^\top)}$. $$ \phi'=\begin{bmatrix} -\sin\alpha\cos\beta&-\cos\alpha\sin\beta\\ \cos\alpha\cos\beta&-\sin\alpha\sin\beta\\ 0&\sin2\beta \end{bmatrix}\\ \phi'^{\top}\phi'=\begin{bmatrix} -\sin\alpha\cos\beta&\cos\alpha\cos\beta&0\\ -\cos\alpha\sin\beta&-\sin\alpha\sin\beta&\sin2\beta \end{bmatrix} \begin{bmatrix} -\sin\alpha\cos\beta&-\cos\alpha\sin\beta\\ \cos\alpha\cos\beta&-\sin\alpha\sin\beta\\ 0&\sin2\beta \end{bmatrix}\\ =\begin{bmatrix} \cos^2\beta&0\\ 0&\sin^2\beta+\sin^22\beta \end{bmatrix}\\ \sqrt{\det(\phi'^\top\phi')}=\sin\beta\cos\beta\sqrt{1+4\cos^2\beta}\\ f(x,y,z)={1\over2}\sqrt{1+4\cos^2\beta} $$ So the required integral is $$\begin{align} &{1\over2}\int_{0}^{2\pi}\int_0^{\pi\over2}\sin\beta\cos\beta(1+4\cos^2\beta)d\beta d\alpha\\ =&\pi\int_0^{\pi\over2}\sin\beta\cos\beta(1+4\cos^2\beta)d\beta\\ =&\pi\int_0^{1}(t+4t^3)dt\quad[\;t=\cos\beta\;]\\ =&{3\pi\over2} \end{align} $$
You could've chosen the Cartesian parametrization of your surface which looks like $$ \phi(x,y)=(x,y,1-x^2-y^2),\quad x^2+y^2<1 $$ In this case the square-root of Gramian determinant turns out to be $\sqrt{1+4x^2+4y^2}$ and therefore our integral becomes $$ {1\over2}\int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}(1+4x^2+4y^2)dxdy\\ ={1\over2}\int_0^{2\pi}\int_0^1(1+4r^2)rdrd\theta\\ ={3\pi\over2} $$