MA 1505 Tutorial 8: Surface Area and Volume

Assume z=z(x,y) is a surface on \mathbb{R}^{3}, the domain R is the projection of the surface z=z(x,y)  on xy-plane. Then the area of the surface is

\iint_{R} \sqrt{1+z_{x}^{2}+z_{y}^{2}} dxdy,

where z_{x} and z_{y} are partial derivatives of z=z(x,y) with respect to the variable x and y respectively.

If a surface is z=z(x,y)\geq 0 and the projection of it on xy-plane is R, then the volume bounded by xy-plane and the surface z=z(x,y) is

\iint_{R} z(x,y) dxdy.

Theorem 1. 

The surface area of the sphere with radius R is 4\pi R^{2}.

The volume of the sphere with radius R is \frac{4}{3} \pi R^{3}.

Proof.  The equation of the sphere with radius R is x^{2}+y^{2}+z^{2}=R^{2}.

First we calculate the surface area of sphere.

Assume R=\{ (x,y): x^{2}+y^{2} \leq R^{2} \}, the function z=z(x,y)=\sqrt{ R^{2}-x^{2}-y^{2}}.

Then

z_{x}=(-x)/ \sqrt{R^{2}-x^{2}-y^{2}},

z_{y}=(-y)/ \sqrt{R^{2}-x^{2}-y^{2}}.

Therefore

\sqrt{1+z_{x}^{2}+z_{y}^{2}} =R/ \sqrt{R^{2}-x^{2}-y^{2}}.

The surface area of half-sphere is

\iint_{x^{2}+y^{2}\leq R^{2}} \frac{R}{\sqrt{R^{2}-x^{2}-y^{2}}} dxdy

= \int_{0}^{2\pi} \int_{0}^{R} \frac{Rr}{\sqrt{ R^{2}-r^{2}}} dr d\theta

= 2 \pi R \int_{0}^{R} \frac{r}{\sqrt{R^{2}-r^{2}}} dr

= 2 \pi R^{2}

Hence, the total surface area of the sphere with radius R is 4\pi R^{2}.

Second, we calculate the volume of the sphere with radius R.

V= 2\iint_{x^{2}+y^{2} \leq R^{2}} \sqrt{R^{2}-x^{2}-y^{2}} dx dy

= 2 \int_{0}^{2\pi} \int_{0}^{R} \sqrt{R^{2}-r^{2}}\cdot r dr d\theta

= 2 \cdot 2\pi \cdot \int_{0}^{R} \sqrt{R^{2}-r^{2}}\cdot r dr

= \frac{4}{3} \pi R^{3}.

Theorem 2. The volume of the ellipsoid \frac{x^{2}}{a^{2}} +\frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}}=1 is \frac{4}{3}\pi abc.

Proof.

The upper bound of the volume is

z= c\cdot \sqrt{1-\frac{x^{2}}{a^{2}}- \frac{y^{2}}{b^{2}}}.

The lower bound of the volume is

z=- c\cdot \sqrt{1-\frac{x^{2}}{a^{2}}- \frac{y^{2}}{b^{2}}}.

Assume R=\{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1 \}, the volume of the ellipsoid is

V= 2 \iint_{R} c \cdot \sqrt{1-\frac{x^{2}}{a^{2}}- \frac{y^{2}}{b^{2}}} dxdy

= 2 c \int_{0}^{2\pi} \int_{0}^{1} \sqrt{1-r^{2}} \cdot (r \cdot a \cdot b) dr d\theta

where we use the substitution x=a \cdot r \cos \theta and y=b\cdot r \sin \theta, the determinant of Jacobian matrix is a\cdot b\cdot r.

Therefore, the value equals to

2abc \cdot (2\pi) \int_{0}^{1} \sqrt{1-r^{2}} r dr = \frac{4}{3} \pi abc.

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