Department of Mathematics Assignment -I1. Find the volume of a solid bounded by the spherical surface 𝑥2 + 𝑦2 +𝑧2 = 4𝑎2 and the cylinder 𝑥2 + 𝑦2 − 2𝑎𝑦 = 0.2. Find the value of 𝑥2 𝜕2𝑢𝜕𝑥2 + 2𝑥𝑦 𝜕2𝑢𝜕𝑥𝜕𝑦 + 𝑦2 𝜕2𝑢𝜕𝑦2 if 𝑢 = sin−1(𝑥3 + 𝑦3)25.3. Find average density of the sphere of radius 𝑎 whose density at a distance𝑟 from the centre of the sphere is 𝜌 = 𝜌0 [1 + 𝑘 𝑟3𝑎3].4. Find by double integration, the volume of the solid bounded under thesurface 𝑓(𝑥, 𝑦) = 𝑦(𝑎−𝑥)√𝑎𝑥−𝑦2 over the region D defined by 𝑦 ≤ 𝑥 ≤ 𝑦2𝑎and 0 ≤ 𝑦 ≤ 𝑎

To find the volume of the solid bounded by the spherical surface 𝑥2 + 𝑦2 +𝑧2 = 4𝑎2 and the cylinder 𝑥2 + 𝑦2 − 2𝑎𝑦 = 0.2, we need to determine the intersection points of these two surfaces.

To find the intersection points, we can substitute the equation of the cylinder into the equation of the sphere. This will give us a quadratic equation in terms of 𝑥 and 𝑧.

Solving this quadratic equation will give us the values of 𝑥 and 𝑧 at the intersection points. We can then substitute these values into the equation of the sphere to find the corresponding values of 𝑦.

Once we have the coordinates of the intersection points, we can use the triple integral to find the volume of the solid. The limits of integration will be determined by the intersection points.

To find the value of 𝑥2 𝜕2𝑢𝜕𝑥2 + 2𝑥𝑦 𝜕2𝑢𝜕𝑥𝜕𝑦 + 𝑦2 𝜕2𝑢𝜕𝑦2, we need to differentiate the function 𝑢 = sin−1(𝑥3 + 𝑦3)25.3 twice with respect to 𝑥 and 𝑦.

After finding the second partial derivatives, we can substitute the values of 𝑥 and 𝑦 into the expression to find the value.

To find the average density of the sphere of radius 𝑎, we need to integrate the density function 𝜌 = 𝜌0 [1 + 𝑘 𝑟3𝑎3] over the volume of the sphere.

The limits of integration will be determined by the radius 𝑎 and the distance 𝑟 from the center of the sphere.

After integrating the density function over the volume of the sphere, we can divide the result by the volume of the sphere to find the average density.

To find the volume of the solid bounded under the surface 𝑓(𝑥, 𝑦) = 𝑦(𝑎−𝑥)√𝑎𝑥−𝑦2 over the region D defined by 𝑦 ≤ 𝑥 ≤ 𝑦2𝑎 and 0 ≤ 𝑦 ≤ 𝑎, we need to perform a double integration.

The limits of integration will be determined by the region D.

After setting up the double integral, we can evaluate it to find the volume of the solid.