Let dv be a infinitely small volume element of a sphere. Now we know in Spherical Polar Coordinate dv=r^2 sin(Z) dr dZ dP where Z is the radial angle and P is the cross radial angle. Now as per problem the radius is limited in a i.e., 0<= r <= a , 0<= Z <= Pi , 0<=P<= 2pi.
Now Volume = Triple Integration of dv with the limit mentioned above = Integration (over limit o to a ) r^2 dr integration (over 0 to pi) sin(z) integration (0 to 2 pi) dP
=> Volume = (a^3/3)*2*2pi = 4/3 pi a*a*a
**For this solution you need to have a working knowledge of Multivariable calculus.
I hope this answer helps. All the very best for your future endeavors!
Question : A hollow sphere has an outer radius of 6 cm and inner radius of 3 cm. What is the volume of this hollow sphere?
Option 1: 252$\pi$ cm3
Option 2: 356$\pi$ cm3
Option 3: 144$\pi$ cm3
Option 4: 175$\pi$ cm3
Question : The height of a cylinder is $\frac{2}{3}$rd of its diameter. Its volume is equal to the volume of a sphere whose radius is 4 cm. What is the curved surface area (in cm2) of the cylinder?
Option 1: $\frac{112}{3} \pi$
Option 2: $32 \pi$
Option 3: $\frac{128}{3} \pi$
Option 4: $40 \pi$
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