求图形面积（1）x=a（cost+tsint）,y=a（sint-tcost）（0≤t≤2pi,a＞0）与x=a（2）r=(3asintcost)/sin^3(t)+cos^3(t)（3）x^2+y^2+z^2≤1与柱体（x-0.5）^2+y^2≤0.25

Question

求图形面积
（1）x=a（cost+tsint）,y=a（sint-tcost）（0≤t≤2pi,a＞0）与x=a
（2）r=(3asintcost) / sin^3(t)+cos^3(t)
（3）x^2+y^2+z^2≤1 与柱体（x-0.5）^2+y^2≤0.25

Answer 1

(1).x=a(cost+tsint),y=a(sint-tcost)；(0≤t≤2π,a＞0)与x=a
这是一条半径为a的圆的渐开线.
ρ²=x²+y²=a²(cost+tsint)²+a²(sint-tcost)²
=a²[(cos²t+2tsintcost+t²sin²t)+(sin²t-2tsintcost+t²cos²t)]
=a²(1+t²)
故所围面积S=[0,2π](1/2)∫ρ²dt=[0,2π](a²/2)∫(1+t²)dt
=(a²/2)(t+t³/3)∣[0,2π]=(a²/2)[2π+(8/3)π²]=πa²[1+(4/3)π].
(2).r=(3asintcost) /(sin³t+cos³t)
r=3a(xy/r²)/[(y³+x³)/r³]=3arxy/(x³+y³),即有x³+y³=3axy；这是一条笛卡儿叶形线；
当0≦t≦π/2时曲线构成一个封闭的图形； 此封闭图形的面积：
S=[0,π/2](1/2)∫r²dt=[0,π/2](9a²/2)∫[(sintcost)/(sin³t+cos³t)]²dt
=[0,π/2](9a²/2)∫[(tantsect)/(1+tan³t)]²dt=[0,π/2](9a²/2)∫[tan²td/(1+tan³t)²]d(tant)
=[0,π/2](3a²/2)∫[d(1+tan³t)/(1+tan³t)²]=(3a²/2)[-1/(1+tan³t)]∣[0,π/2]=3a²/2.
(3).x²+y²+z²≤1 与柱体[x-(1/2)]²+y²≤1/4
z=±√(1-x²-y²)基于对称性取z=√(1-x²-y²)；
∂z/∂x=-x/√(1-x²-y²)；∂z/∂y=-y/√(1-x²-y²)；1+(∂z/∂x)²+(∂z/∂y)²=1/(1-x²-y²)
面积S=[D] 2∫∫√[1+(∂z/∂x)²+(∂z/∂y)²]dxdy=[D] 2∫∫dxdy/√(1-x²-y²)
区域Dxy:[x-(1/2)]²+y²≤1/4； 用极坐标：x=rcosθ,y=rsinθ；-π/2≦θ≦π/2；0≦r≦cosθ
故S=[-π/2,π/2]2∫dθ[0,cosθ]∫rdr/√(1-r²)=[-π/2,π/2]2∫dθ[0,cosθ](-1/2)∫d(1-r²)/√(1-r²)
=[-π/2,π/2]2∫dθ[0,cosθ](-1/2)[2√(1-r²)][0,cosθ]
=[-π/2,π/2](-2)∫(sinθ-1)dθ=2(cosθ+θ)∣[-π/2,π/2]=2(π/2+π/2)=2π.