Answer
$4 \pi b (b-\sqrt {b^2-a^2})$
Work Step by Step
$Surface \ Area ; A(S)=\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial y} )^2 } dA= \iint_{D} \sqrt {1+(\dfrac{-x}{\sqrt {b^2-x^2-y^2}})^2+(\dfrac{-y}{\sqrt {b^2-x^2-y^2}})^2} dA \\= \iint_{D} \dfrac{b}{\sqrt {b^2-x^2-y^2}} dA \\=\int_{-a}^{a} \int_{-\sqrt{a^2-x^2}}^{\sqrt {a^2-x^2}} \dfrac{b}{\sqrt {b^2-x^2-y^2}} dx dy \\ = \int_{0}^{2 \pi} \int_{0}^{a} \dfrac{b}{\sqrt {b^2-r^2}} \ r \ dr \ d \theta \\ = b \int_0^a \dfrac{r}{\sqrt {b^2-r^2}} dr \times \int_0^{2 \pi} d \theta$
or, $=2 \pi b [-\sqrt{b^2-a^2} -(-b)]$
or, $=2 \pi b (b-\sqrt {b^2-a^2})$
Now, the total surface area of the part of the sphere $x^2+y^2+z^2=b^2$ that lies inside the cylinder becomes: $ 2\times [2 \pi b (b-\sqrt {b^2-a^2})] =4 \pi b (b-\sqrt {b^2-a^2})$
