Answer
$$4 \pi b (b-\sqrt {b^2-a^2})$$
Work Step by Step
We have $A(S)=\iint_{D} \sqrt {1+(z_x )^2+(z_y )^2 } dA \\= \iint_{D} \sqrt {1+(\dfrac{-x}{\sqrt {b^2-x^2-y^2}})^2+(\dfrac{-y}{\sqrt {b^2-x^2-y^2}})^2} dA \\= \iint_{D} \dfrac{b}{\sqrt {b^2-x^2-y^2}} dA $
and $\iint_{D} dA$ is the area of the region $D$
Now, $A(S)=\int_{-a}^{a} \int_{-\sqrt{a^2-x^2}}^{\sqrt {a^2-x^2}} \dfrac{b}{\sqrt {b^2-x^2-y^2}} dx dy \\= \int_{0}^{2 \pi} \int_{0}^{a} \dfrac{b}{\sqrt {b^2-r^2}} r dr d \theta \\= b \int_0^a \dfrac{r}{\sqrt {b^2-r^2}} dr \int_0^{2 \pi} d \theta \\= b [-\sqrt{b^2-a^2} -(-b)] \times 2 \pi \\=2 \pi b (b-\sqrt {b^2-a^2})$
Now, the total surface area of the part of the sphere is equal to $x^2+y^2+z^2=b^2$ that lies inside the cylinder is:
$ 2 \times [2 \pi b (b-\sqrt {b^2-a^2})] =4 \pi b (b-\sqrt {b^2-a^2})$
