Question

高数A 求面积

求解

Answer 1

5. x=y²/2=y+4
所以y=4,y=-2
所以面积=∫(-2,4)[(y+4)-y²/2]dy
=(y²/2+4y-y³/6) (-2,4)
=(8+16-64/6)-(2-8+8/6)
=18

追答

6. 这种积分题还是比较麻烦的，真想用matlab给你做。这是个“鸡蛋图”
只求y大于0部分的面积，记为s1
极坐标化为参数方程：x=2a(2+cost)cost，y=2a(2+cost)sint
s1=int(π/2,0)(2a(2+cost)sint)d(2a(2+cost)cost)
=(-8a^2)int(π/2,0)((2sint+sintcost)(sint+sintcost))dt
记积分号里面的为k1=(2sint+sintcost)(sint+sintcost)=2sint^2+3sint^2cost+sint^2cost^2
记s11=int(π/2,0)(2sint^2)dt=(t-sin2t/2)|(π/2,0)=-π/2
s12=int(π/2,0)(3sint^2cost)dt=sint^3|(π/2,0)=-1
s13=int(π/2,0)(sint^2cost^2)dt=(1/4)int(π/2,0)(1-cos2t^2)dt=(1/8)int(π/2,0)(1-cos4t)dt
=(1/8)(t-sin4t/4)|(π/2,0)=-π/16
所以s1=(-8a^2)*(-π/2-1-π/16)=(9π+16)a^2/2
所求面积为s1的2倍，即s=2s1=(9π+16)a^2