Answer to Question #246036 in Calculus for Njabulo

Question

8. Evaluate the line integral
Z
C
(xy + z) ds
where C is the line segment in R
3 with initial point (1, 0 − 1) and endpoint (2, 1, 1). [8]
9. Let D be the region in R
3
that lies inside the sphere x
2 + y
2 + z
2 = 2 and above the plane z = 1.
(a) Sketch the region D in R
3
. (3)
(b) Express the volume of D in terms of a triple integral, using spherical coordinates. DO NOT
EVALUATE THE INTEGRAL. (7)
[10]
10. Let S be that part of the surface z =
p
x
2 + y
2 that lies between the plane z = 1 and the plane z = 3.
(a) Sketch the surface S, together with its XY-projection. (3)
(b) Use a surface integral to determine the area of S

Answer 1

8.

Parametrise the line:

$r(t)=(1,0,-1)+t[(2,1,1)-(1,0,-1)]=(t+1,t,2t-1), t\isin [0,1]$

Then:

$r'(t)=(1,1,2)$

$|r'(t)|=\sqrt{1+1+4}=\sqrt{6}$

$ds=|r'(t)|dt=\sqrt{6}dt$

$f(r(t))=t^2+3t-1$

$\int(xy+z)ds=\sqrt{6}\int^1_0(t^2+3t-1)dt=\sqrt{6}(t^3/3+3t^2/2-t)|^1_0=5/\sqrt{6}$

9.

a)

b)

maximum value of $\phi$ :

$cos\phi=1/\sqrt{2}$

$\phi=\pi/4$

minimum value of $\phi$ :

$cos\phi=1/\rho$

$\rho=sec\phi$

$V=\iiint dV=\int^{2\pi}_{0}\int^{\pi/4}_{0}\int^{\sqrt{2}}_{sec\phi}\rho^2sec\phi d\rho d\phi d\theta$

10.

a)

The XY -projection:

b)

$f(x,y)=\sqrt{x^2+y^2}$

$\frac{df}{dx}=\frac{x}{\sqrt{x^2+y^2}}$

$\frac{df}{dy}=\frac{y}{\sqrt{x^2+y^2}}$

$\sqrt{(\frac{df}{dx})^2+(\frac{df}{dx})^2+1}=\sqrt{2}$

$\iint ds=\iint \sqrt{2}dR$

 in polar co-ordinates:

$R=\{(r,\theta): 1\le r\le3,0\le \theta\le 2\pi\}$

$\iint \sqrt{2}dR=\int^{2\pi}_0\int^3_1 \sqrt{2}rdrd\theta=\frac{\sqrt{2}}{2}\int^{2\pi}_0 r^2|^3_1d\theta=4\sqrt{2}\int^{2\pi}_0 d\theta=8\pi\sqrt{2}$