# 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

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