The volume of the sphere x2+y2+z2=1𝑥2+𝑦2+𝑧2=1 is evaluated by:a.4∫10∫1−y2√01−x2−y2−−−−−−−−−√dxdy4∫01∫01−𝑦21−𝑥2−𝑦2𝑑𝑥𝑑𝑦b.8∫10∫1−y2√01−x2−y2−−−−−−−−−√dxdy8∫01∫01−𝑦21−𝑥2−𝑦2𝑑𝑥𝑑𝑦c.8∫10∫101−x2−y2−−−−−−−−−√dxdy8∫01∫011−𝑥2−𝑦2𝑑𝑥𝑑𝑦d.16∫10∫1−y2√01−x2−y2−−−−−−−−−√dxdy

The volume of a sphere with radius r is given by the formula V = 4/3πr³. In this case, the equation of the sphere is x² + y² + z² = 1, which means the radius of the sphere is 1. Therefore, the volume of the sphere is V = 4/3π(1)³ = 4/3π.

The integrals given in the options are in spherical coordinates. The volume element in spherical coordinates is given by dV = r²sin(θ)drdθdφ.

The limits of the integrals suggest that we are integrating over the entire sphere, which would give us the volume of the sphere.

However, none of the integrals given in the options match the volume of the sphere calculated using the formula V = 4/3πr³. Therefore, without further information, it is not possible to determine which of the options is correct.