Question

急！我是数学白痴，问一道简单的二重积分题，在线等！

如图，请知道的朋友帮忙写一下解题过程，谢谢！

Answer 1

解：原式=∫(0,1)dx∫(0,x²)√(1+x²)dy (交换积分顺序)
=∫(0,1)[x²√(1+x²)]dx
=∫(0,π/4)(tan²t*sect*sec²t)dt (设x=tant)
=∫(0,π/4)[sin²t/(cos²t)³]d(sint)
=∫(0,π/4)[sin²t/(1-sin²t)³]d(sint)
=1/16∫(0,π/4)[(3sint-sin²t)/(1-sint)³-(3sint+sin²t)/(1+sint)³]d(sinx)
=1/48∫(0,π/4){[(-3sin²t+6sint-3)-3(1-sint)+6]/(1-sint)³
-[(3sin²t+6sint+3)+3(sint+1)-6]/(1+sint)³}d(sint)
=1/48{∫(0,π/4)d[(1-sint)³]/(1-sint)³+3∫(0,π/4)d(1-sint)/(1-sint)²
-6∫(0,π/4)d(1-sint)/(1-sint)³-∫(0,π/4)d[(1+sint)³]/(1+sint)³
-3∫(0,π/4)d(1+sint)/(1+sint)²+6∫(0,π/4)d(1+sint)/(1+sint)³}
=1/48[3ln(1-sint)-3/(1-sint)+3/(1-sint)²-3ln(1+sint)+3/(1+sint)-3/(1+sint)²]│(0,π/4)
=1/48{3ln[(2-√2)/(2+√2)]-6/(2-√2)+12/(2-√2)²+6/(2+√2)-12/(2+√2)²}
=1/48[6ln(√2-1)-6√2+24√2]
=1/8[ln(√2-1)+3√2]

Answer 2

等下哈！偶做好放图片上偶的空间再发连接给你哦！
有点麻烦的一道题目！！
等偶做了图片你看了之后，满意了再选偶吧！！
等10分钟左右哦！
不要妄自菲薄啊！
呵呵！偶重修了5门！这些题目偶还不是一样觉得很轻松！
不会做一些题目算不上白痴啊！呵呵！

http://hi.baidu.com/%C1%E3%CF%C2%B8%BA5%B6%C8%D0%A1/album/item/f5d7341452272d34c93d6dcf.html

参考资料
http://hi.baidu.com/%C1%E3%CF%C2%B8%BA5%B6%C8%D0%A1/album/item/f5d7341452272d34c93d6dcf.html#IMG=11f87f25c67eac2b8b82a170

崩溃了！！！
算错啦！！....！！不想算了！太痛苦了....1...
方法类似的！！
如果你能找到不用算得像我那么扑街的算法！记得告诉我啊！
等高手吧你！！！

Answer 3

算不上白痴,极难!
∫(0,1)dy ∫(y^(1/2),1) (1+x^2)^(1/2) dx
=∫(0,1)(1-x^2)^(1/2) dx ∫(0,x^2) dy
=∫(0,1)(x^2)*(1-x^2)^(1/2) dx
=∫(0,1) {(1-x^2)^(1/2)}/4[3x^2-(1-x^2)+1]dx
=∫(0,1) {(3/4)x^2(1-x^2)^(1/2)-(1/4)(1-x^2)^(3/2)
+(1-x^2)^(1/2)}dx
=∫(0,1)(x/4)(3/2)(1-x^2)^(1/2)(-2x)-(1/4)(1-x^2)^(3/2)
+(1/8)((1-x^2)+x(1/2)(1-x^2)^(-1/2) (-2x)+(1/8)(1-x^2)^(-1/2)}dx
={(-x/4)*(1-x^2)^(3/2)
+(1/8)(x(1-x^2)^(1/2)+arcsinx)}(x=0,1)
=п/16