计算二重积分$$ \int \int _ { D } \int ( x + 2 y ) d x d y $$其中D是由直线$$ x \neq 0 $$$ y = 0 $$以及$$ x + y = 1 $$所围成的闭区域。 相关知识点: 试题来源: 解析 七、1.我以日始出时去人近,而日中时远也 日初出远 日中时近 远者小而近者大乎...
【题目】计算二重积分(x+y)dxdy,其中D为x^2+y^2≤2x 答案 【解析】 楼上错的,楼上当作矩形区域算了 首先本题区域关于x轴对称,y关于y是一个奇函 数,因此积分为0,所以被积函数中的y可去掉。 J∫(x+y)dxdy =∫rdxdy 用极坐标,x2+y2=2x的极坐标方程为:=2c0s0 =J[-T/2---...
Explain why ∫−∞∞∫−∞∞e−(x2+y2)dxdy=(∫−∞∞e−x2)2. Double Integrals and Solving Them: Let D⊂R2 be some area we want to integrate our function over. Say that function is f:D→R. The result of the integration represents th...
{eq}\;\iint\limits_D {f\left( {x,y} \right)}dxdy = \iint\limits_{{D^*}} {f\left( {r\cos \theta ,rsin\;\theta } \right)rdrd\theta }. {/eq} Answer and Explanation: Taking into account the region is delimited by a circl...
先积y, ff$$ x ^ { 2 } $$dxdy $$ = \int \left[ - 1 - \cdots > 3 \right] d x \int \left[ - \sqrt { ( 3 - x ^ { 2 } + 2 x ) } - \cdots - - - - - - - - \sqrt { ( 3 - x ^ { 2 } + 2 } $$ x)] $$ x ^ { 2 } $$|dy $$ ...
解析 【解析】 $$ \frac { 3 2 } { 9 } $$; 结果一 题目 【题目】计算二重积分∫_b^x√(x^2+y^2)dxdy ,其中 D:x^2+y^2≤x 答案 【解析】积分区域D如图(省略),则积分区域表示为-x/2≤0≤x/2;0≤x≤cosθ. 于是∫_0^1√(x^2+y^2)dxdy=∫_(-π/2)^(π/...
解析 取z=0下侧为∑1z=3上侧为∑2那么∫∫∑1 xdydz+ydzdx+zdxdy=0 ∫∫∑2 xdydz+ydzdx+zdxdy=3∫∫dxdy=3(9π)=27π且根据高斯公式∫∫∑+∑1+∑2 xdydz+ydzdx+zdxdy=3∫∫∫dV=3(3x9π)=81π所以原积分=81π-0-27π=54π 反馈 收藏 ...
Answer to: Evaluate the following double integral. \int_{ - 1}^3 {\int_{{y^2}}^{2y + 3} {\left( {x + y} \right)dxdy} } By signing up, you'll get...
Answer to: Evaluate \int\int f(x,y)dxdy , when , f(x,y)= (yx)^3 and , R= [0,2] x [-1,0]. By signing up, you'll get thousands of step-by-step...
计算∫∫(x+y)^2dxdy,区域D由x^2+y^2≥2x,x^2+y^2≤4x所围 答案 D关于轴对称→yd=aydo=0 (+do=(+2ry+y)do +y )do=2 o=d0 = =(256cos0-16cos0)d0 =120cos40d0=120.31、=120.3-1.-45 4!2 4·222相关推荐 1计算∫∫(x+y)^2dxdy,区域D由x^2+y^2≥2x,x^...