Answer to Question #218913 in Calculus for Nkululeko

Question

Consider the surface: S={(x,y,z)|z=sqr(x^2+y^2) and 1</=z</=3}. (a) sketch the surface S in R^3. Also show its XY-projection on your sketch. (b) evaluate the area of S, using a surface integral.

Answer 1

ANSWER.

Denote $D$ the XY-projection of surface $S$. $D=\ \left\{ \left( x,y \right) :1\le { x }^{ 2 }+{ y }^{ 2 }\le 9 \right\}$.

Since $S=\left\{ \left( x,y,f(x,y) \right) :\left( x,y \right) \in D,f(x,y)=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \right\}$ then to calculate area of $S$ we have $A=\iint _{ S }^{ \ }{ dS\ =\iint _{ D }^{ \ \ }{ \sqrt { 1+{ \left( \frac { \partial f }{ \partial x } \right) }^{ 2 }+{ \left( \frac { \partial f }{ \partial y } \right) }^{ 2 } } dxdy } }$ .

$\sqrt { 1+{ \left( \frac { \partial f }{ \partial x } \right) }^{ 2 }+{ \left( \frac { \partial f }{ \partial y } \right) }^{ 2 } } =\sqrt { 2 }$ , because $\frac { \partial f }{ \partial x } =\frac { x }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } } ,\frac { \partial f }{ \partial y } =\frac { y }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } }$ . Therefore

$A=\iint _{ D }^{ \ }{ \sqrt { 2^{ \quad } } dxdy } =\sqrt { 2^{ \ } } \iint _{ D }^{ \ }{ \ dxdy }$ . To calculate this integral ,replace Cartesian coordinates with polar coordinates : $A=\quad \sqrt { 2^{ \ } } \int _{ 0 }^{ 2\pi }{ \int _{ 1 }^{ 3 }{ rdrd\theta } } =2\sqrt { 2 } \pi \int _{ 1 }^{ 3 }{ rdr=\sqrt { 2 } \pi { \left( { r }^{ 2 } \right) }_{ 1 }^{ 3 }=\sqrt { 2 } \pi (9-1)=8\sqrt { 2 } \pi \quad }$.