269 int_(0)^1dyint_(y)^1sqrt(x^2)-y^(2)dx的值为A. $\frac{\pi}{3}$.B. $\frac{\pi}{6}
17.计算int_(1)^2dyint_(sqrt(y-1))^1(sin x)/(x)dx.17.计算$\int_{1}^{2}dy\int_{\sqrt{y
函数 y=sqrt(2-x^2-y^2)+(1)/(sqrt(x^2+y^2-1)) 的定义域是A. $\{(x,y)| 1 \leq x^2 + y^2 \l
267 累次积分 int_(-(pi)/(2))^(pi)/(2)dxint_(0)^sin x(x^2+ycosx)sqrt(1-y^2)dy=A. $\fr
int_(0)^2(1)/(1+sqrt[3](x))dx;$\int_{0}^{2}\frac{1}{1+\sqrt[3]{x}}dx;$
6 已知平面区域D=((x,y)|sqrt(1-y^2)le xle 1,-1le yle 1),计算iintlimits_(D)(x)/(sqrt(x^2)+
求下列函数的自然定义域:(1)y=sqrt(3x+2);(2)y=dfrac(1)(1-{x)^2};(3)y=dfrac(1)(x)-sqrt(1-(x)^2
计算(int )_(0)^1dx(int )_(1-x)^sqrt (1-{x^2)}dfrac (x+y)({x)^2+(y)^2}dy=-|||-dv=__
设 y = y ( x ) 满足 +dfrac (1)(2sqrt {x)}y=2+sqrt (x), y(1)=3,求 y = y ( x ) 的渐近 线设y
(6) int_((1)/(sqrt(2)))^1(sqrt(1-x^2))/(x^2)dx;(6) $\int_{\frac{1}{\sqrt{2}}}^{1