Answer
$$V = \pi $$
Work Step by Step
$$\eqalign{
& y = {\left( {1 - {x^2}} \right)^{ - 1/2}}{\text{ and }}y = 0{\text{ over the interval }}\left[ {0,\frac{{\sqrt 3 }}{2}} \right] \cr
& {\text{Revolved about }}y{\text{ - axis}} \cr
& {\text{From the graph of the region shown below}} \cr
& {\left( {1 - {x^2}} \right)^{ - 1/2}} > 0{\text{ on the interval }}\left[ {0,\frac{{\sqrt 3 }}{2}} \right] \cr
& {\text{Using the shell method about the }}y{\text{ - axis}} \cr
& V = \int_a^b {2\pi x\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{Let }}f\left( x \right) = {\left( {1 - {x^2}} \right)^{ - 1/2}}{\text{ and }}g\left( x \right) = 0 \cr
& V = \int_0^{\sqrt 3 /2} {2\pi x\left[ {{{\left( {1 - {x^2}} \right)}^{ - 1/2}}} \right]} dx \cr
& V = - \pi \int_0^{\sqrt 3 /2} {{{\left( {1 - {x^2}} \right)}^{ - 1/2}}\left( { - 2x} \right)} dx \cr
& {\text{Integrating}} \cr
& V = - 2\pi \left[ {{{\left( {1 - {x^2}} \right)}^{1/2}}} \right]_0^{\sqrt 3 /2} \cr
& V = - 2\pi \left[ {{{\left( {1 - {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \right)}^{1/2}} - {{\left( {1 - {{\left( 0 \right)}^2}} \right)}^{1/2}}} \right] \cr
& V = - 2\pi \left[ {{{\left( {\frac{1}{4}} \right)}^{1/2}} - {{\left( 1 \right)}^{1/2}}} \right] \cr
& V = - 2\pi \left( {\frac{1}{2} - 1} \right) \cr
& {\text{Simplifying}} \cr
& V = \pi \cr} $$
