计算I=∫∫√1-x2-y2/1 x2 y2dxdy,其中D=
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假设x^2+y^2=m那么m(m+1)=20即(m+5)(m-4)=0那么m=-5或4所以x^2+y^2=4
用极坐标来解吧,令x=r*cosθ,y=r*sinθ那么显然√(x²+y²)=r,由x²+y²≤2x可以得到r²≤2r*cosθ即r≤2cosθ故r的
由于1=x2+y2+z2=(x2+12y2)+(12y2+z2)≥2x•y2+2•y2•z=2(xy+yz),当且仅当x=y2=z时,等号成立,∴x=y2=z=12时,xy+yz的最大值为22.故答案
(x-1)^2+(y-1)^2=1令x-1=sinay-1=cosa则x=1+sina,y=1+cosax^2+y^2=1+2sina+(sina)^2+1+2cosa+(cosa)^2=3+2(si
设x2+y2=t,则方程即可变形为t(t-1)-12=0,整理,得(t-4)(t+3)=0,解得t=4或t=-3(不合题意,舍去).即x2+y2=4.
1,16x2+9y2=144化为标准方程x²/(144/16)+y²/(144/9)=1x²/(12/4)²+y²/(12/3)²=1x&s
∵(x2+y2+1)2-4=0,∴(x2+y2+1)2=4,∵x2+y2+1>0,∴x2+y2+1=2,∴x2+y2=1.故答案为:1.
是一个高为1的碗形旋转抛物面,底圆半径为1,转换成极坐标,V=4∫[0,π/2]dθ∫[0,1][(rcosθ)^2+(rsinθ)^2]rdr=4∫[0,π/2]dθ∫[0,1]r^3dr=4∫[0
dS=√(1+4x^2+4y^2)dxdy,投影:x^2+y^2《1I=∫∫1/(x^2+y^2+(x^2+y^2)^2)*√(1+4x^2+4y^2)dxdy+∫∫1/(x^2+y^2+1)*dxd
R=x^2zRz=x^2由高斯公式:I=∫∫x2zdxdy=∫∫∫x^2dxdydz(xoy平面的投影D:x^2+y^2
(x+y)^2=1+3xy(x-y)^2=1-xyu=(x+y)(x-y)|u|=√(x+y)^2√(x-y)^2=√(1+3xy)√(1-xy)=√[-3(t-1/3)^2+2/3]≤√6/3故-√
(x2-2x+1-y2)÷(x+y-1),=(x-1+y)(x-1-y)÷(x+y-1),=x-y-1.故应填:x-y-1.
1设Z=cos(xy2)+3x/x2+y2,计算δz/δyδz/δy=-2xy*sin(xy2)-(3x*2y)/(x2+y2)22、设Z=f(x2-y2,exy),其中f(u,v)为可微函数,求dz
∵x2+xy=5,xy+y2=-1,∴(x2+xy)-(xy+y2)=x2+xy-xy-y2=x2-y2=5-(-1)=6.故填:6
可设x²+y²=t.则t(t-1)=2.===>t²-t-2=0.===>(t-2)(t+1)=0.===>t=2.即x²+y²=2.
先两组基之间的过渡矩阵P(y1,y2,y3)=(x1,x2,x3)P则T(x1,x2,x3)=(y1,y2,y3)=(x1,x2,x3)PT(y1,y2,y3)=T(x1,x2,x3)P=(y1,y2
x²+y²/2=1则:2x²+y²=2则:x√(1+y²)=(√2/2)×√[(2x²)(1+y²)]而:√[(2x²)