Question

a) Let Φ(u, v) = (2 sin u cos v, 3 sin u sin v, cos u), for π/2 ≤ u, v ≤ π. Plot and describe the surface S parametrized by Φ.

b) Find the equation of the tangent plane to S at (u, v) = (3π/4, 3π/4).

c) Find the volume of the region in R3 bounded by the surface S and the coordinate planes.

Answer 1

ANSWER

a)The surface S parametrized by Φ is part of the ellipsoid $\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } +{ z }^{ 2 }=1$ , since $\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } +{ z }^{ 2 }=$ $\sin ^{ 2 }{ u } \cos ^{ 2 }{ v+ } \ \sin ^{ 2 }{ u\sin ^{ 2 }{ v+\cos ^{ 2 }{ u\ \ = } \ } }$

$=\sin ^{ 2 }{ u } \left( \cos ^{ 2 }{ v+ } \sin ^{ 2 }{ v } \ \right) +\cos ^{ 2 }{ u=\sin ^{ 2 }{ u } +\cos ^{ 2 }{ u=1 } }$

b)The point on the surface S corresponding to (u, v) = (3π/4, 3π/4) has Cartesian coordinates $(-1,3/2,-1/\sqrt { 2 } )$ . Differentiating the equality $\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } +{ z }^{ 2 }=1\quad$with respect to variables $x$ and $y$ we have: ${ \left( \frac { x }{ 2 } \right) }+2z{ z }_{ x }^{ ' }=0\quad ,\quad 2{ \left( \frac { y }{ 9 } \right) }+2z{ z }_{ y }^{ ' }=0$ .

So, ${ z }_{ x }^{ ' }(-1,3/2)=\frac { \ -1/2 }{ 2/\sqrt { 2 } } =-\ \sqrt { 2 } /4\, \quad { z }_{ y }^{ ' }(-1,3/2)=\frac { 1/6 }{ \ \quad 1/\sqrt { 2 } } =\frac { \ \sqrt { 2 } }{ 6 }$ . Consequently, the vector $\overrightarrow { n } =\left< -\frac { \sqrt { 2 } }{ 4 } ,\frac { \ \sqrt { 2 } }{ 6 } ,-1 \right>$ is a normal vector to the S at the point

$(-1,3/2,-1/\sqrt { 2 } )$ . Therefore, the equation of the tangent plane has the form:

$-\frac { \sqrt { 2 } }{ 4 } \left( x+1 \right) +\frac { \ \sqrt { 2 } }{ 6 } \left( y-\frac { 3 }{ 2 } \right) -\left( z+\frac { 1 }{ \sqrt { 2 } } \right) =0$ or

$z=\frac { -\sqrt { 2 } x }{ 4 } +\frac { \ \sqrt { 2 } y }{ 3 } -\ \sqrt { 2 }$

c) The region to be calculated is 1/8 of the ellipsoid . The volume of the ellipsoid $\frac { { x }^{ 2 } }{ \ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } +\frac { { z }^{ 2 } }{ { c }^{ 2 } } =1$ is $\frac { 4 }{ 3 } \pi abc$ .Therefore, the volume is $\frac { 1 }{ 8 } \cdot \frac { 4 }{ 3 } \pi 2\cdot 3\cdot 1=\pi .$

Calculate the volume using the integral.$Volume=\ \iiint _{ V }^{ \ }{ dzdxdy } =\iint _{ D }^{ \ }{ \left( \sqrt { 1-\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 9 } } \right) dxdy }$ , where $D=\left\{ (x,y):\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } \le 1,\ \right\} \ \cap \ \left\{ (x,y):-2\le x\le 0,\ 0\le y\le 3 \right\}$. Passing to new variables $(r,v)$ $x=2r\cos { v } \ ,\ y=3r\sin { v }$ we get :$\Delta =\left\{ (r,v):0\le r\le 1,\quad \frac { \pi }{ 2 } \le v\le \pi \quad \right\}$ ,

$\iint _{ D }^{ \ }{ \left( \sqrt { 1-\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 9 } } \right) dxdy } =\iint _{ \Delta }^{ \ }{ \sqrt { 1-{ r }^{ 2 } } 6r } drdv=$

$=6\int _{ \frac { \pi }{ 2 } }^{ \pi }{ dv\left( \int _{ 0 }^{ 1 }{ \sqrt { 1-{ r }^{ 2 } } \ r } dr \right) } =\ \frac { 3\pi }{ 2 } \ \left( F(1)-F(0) \right) =\ \pi$ , $(F(r)=\ { -\frac { 2 }{ 3 } \left( 1-{ r }^{ 2 } \right) }^{ 3/2 })$ .