Question #246036
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
1
Expert's answer
2021-10-04T16:12:16-0400

8.

Parametrise the line:

r(t)=(1,0,1)+t[(2,1,1)(1,0,1)]=(t+1,t,2t1),t[0,1]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)=(1,1,2)

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

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

f(r(t))=t2+3t1f(r(t))=t^2+3t-1


(xy+z)ds=601(t2+3t1)dt=6(t3/3+3t2/2t)01=5/6\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 ϕ\phi :

cosϕ=1/2cos\phi=1/\sqrt{2}

ϕ=π/4\phi=\pi/4



minimum value of ϕ\phi :

cosϕ=1/ρcos\phi=1/\rho

ρ=secϕ\rho=sec\phi


V=dV=02π0π/4secϕ2ρ2secϕdρdϕdθV=\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)=x2+y2f(x,y)=\sqrt{x^2+y^2}

dfdx=xx2+y2\frac{df}{dx}=\frac{x}{\sqrt{x^2+y^2}}

dfdy=yx2+y2\frac{df}{dy}=\frac{y}{\sqrt{x^2+y^2}}

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


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


 in polar co-ordinates:

R={(r,θ):1r3,0θ2π}R=\{(r,\theta): 1\le r\le3,0\le \theta\le 2\pi\}


2dR=02π132rdrdθ=2202πr213dθ=4202πdθ=8π2\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}


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