(0,1)为顶点的三角形区域的正向边界:(4)$$ I = \oint _ { L } ( 1 + y ^ { 2 } ) d x + y d $$,其中L为[0,π]上正弦曲线$$ y = \sin $$与$$ y 其中L为椭圆$$ \fr a c { x ^ { 2 } } { a ^ { 2 } } + \fr a c { y ^ { 2 } } { b ^ ...
3. 按两种不同次序将二重积分 $$ \int _ { D } $$f(x,y)dxdy化为二次积分,其中D为:(1)由直线$$ y = x $$及抛物线$$ y ^ { 2 } = 4 x $$所围成的闭区域;(2)由直线$$ y = x , x = 2 $$ 及双曲线$$ y = \frac { 1 } { x } ( x > 0 ) $$所围成的闭...
\int_{0}^{1}\int_{x}^{e^{x 3xy^{2} dy dx Calculate the following iterated integral: integral_{0}^{1} integral_{0}^{1} xy sqrt(x^2 + y^2) dy dx. Calculate the iterated integral: integral_{1}^{4} integral_{1}^{2} (x / y + y / x) dy dx....
evaluate the double integral { \iint x^3y^2dA } and D = (x,y) ,{ 0 \leq x \leq 2,-x \leq y \leq x } { \iint e^{y^2}dA } and D = (x,y) ,{ 0 \leq y \leq 1,0 \leq x \leq y } { \iint (2x + 4y + Evaluate...
于是∫_0^1√(x^2+y^2)dxdy=∫_(-π/2)^(π/2)dθ∫_0^∞r^2dr(4.5t)=1/( =1/3∫_-π/2(1-sin^2θ)dsinθ=1/3(sinθ-1/3sin^3θ)]_-4/3πrcos 结果二 题目 【题目】计算二重积分∫_0^x√(x^2+y^2)dxdy ,其中 ⊃:x^2+y^2≤x 答案 【解析】...
$$ \int f ( x + y ) d x d y \\ = \int \int x d x d y $$ $$用极坐标, x ^ { 2 } + y ^ { 2 } = 2 x 的极坐标方程为: r = 2 \cos \theta \\ = \int \left[ - \pi / 2 - \cdots - > \pi / 2 \right] d \theta \int \left[ 0 - ...
( x y + 1 ) d x = 2 \pi $$ 解法三 利用对称性$$ I = \int \int _ { 0 } ^ { x } x y d x d y + \int _ { 0 } ^ { x } d x d y $$.因为积分域D关于x轴对称,且函数 $$ ( x , y ) = x y $$关于x是奇函数,所以$$ \int \int x y d...
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...
Integrate: \int x^xdx Integrate: \int \frac{e^x}{2e^x + 3e^{-xdx Integrate: \int \frac{e^{2x{(e^{4x}+1)} Integrate \int \sqrt {(x-1)} x^2 dx Integrate: \int \frac{3x^2}{\sqrt{7+x^2dx Integrate \int (r^2-2)(r)(\sqrt{4r^2+1}) dr ...
^2)^x(1+xy^2dxdy=∫_0^xdx =4π+0=4π(2)按原点对称计算.第一个积分是积分区域的面积;对于第二个二重积分,由于f(-x,-y)=(-x)(-y)^2=-xy^2=-f(x,y) 且积分区域关于原点对称,故∫_0^x1/x^2dxdy=0 所以I=∫_(x^2)^1(1+xy^2)dxdy=∫_x^ydx+∫_y^x(x^2dx)dx=4...