题目
将直角坐标系中的二次积分I=int_(0)^1dyint_(y)^2-yf(x,y)dx化为极坐标系中的二次积分，则I=()A. I=int_(0)^(pi)/(4)dthetaint_(0)^(2)/(sintheta+costheta)rho f(rhocostheta,rhosintheta)drhoB. I=int_(0)^(pi)/(4)dthetaint_(0)^(2)/(sintheta+costheta)f(rhocostheta,rhosintheta)drhoC. I=int_((pi)/(4))^(pi)/(2)dthetaint_(0)^(2)/(sintheta+costheta)rho f(rhocostheta,rhosintheta)drhoD. I=int_(0)^(pi)/(4)dthetaint_(0)^(2)/(sintheta)rho f(rhocostheta,rhosintheta)drho
将直角坐标系中的二次积分$I=\int_{0}^{1}dy\int_{y}^{2-y}f(x,y)dx$化为极坐标系中的二次积分，则$I=$()
A. $I=\int_{0}^{\frac{\pi}{4}}d\theta\int_{0}^{\frac{2}{\sin\theta+\cos\theta}}\rho f(\rho\cos\theta,\rho\sin\theta)d\rho$
B. $I=\int_{0}^{\frac{\pi}{4}}d\theta\int_{0}^{\frac{2}{\sin\theta+\cos\theta}}f(\rho\cos\theta,\rho\sin\theta)d\rho$
C. $I=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}d\theta\int_{0}^{\frac{2}{\sin\theta+\cos\theta}}\rho f(\rho\cos\theta,\rho\sin\theta)d\rho$
D. $I=\int_{0}^{\frac{\pi}{4}}d\theta\int_{0}^{\frac{2}{\sin\theta}}\rho f(\rho\cos\theta,\rho\sin\theta)d\rho$
题目解答
答案
A. $I=\int_{0}^{\frac{\pi}{4}}d\theta\int_{0}^{\frac{2}{\sin\theta+\cos\theta}}\rho f(\rho\cos\theta,\rho\sin\theta)d\rho$