✔ 最佳答案
（１）A Solid is enclosed by the surfaces 0<=x, y<=a, x^2+y^2<=a^2 and z=x^2+y^2. Calculate the total surface area of the solid.x,y互換原函數不變 => x,y對稱可以簡化成2d投影到x=0平面來看: -a <= y <= az = y^2 => z' = 2y => 拋物線開口向上A =∫2πy*ds=∫2πy√(1 + z'^2)dy=∫2πy√(1 + 4y^2)dy=∫2π√(1+4y^2)d(1+4y^2)/8= (π/4)*(2/3)*(1+4y^2)^1.5.....y=0~a= (π/6)*[(1 + 4a^2)^1.5 - 1]
（２）Let A and B be distince positive number. Calculate∫(0→2pi)[dt／A^2cos^2(t) + B^2sin^2(t)], B < A
Basic formulae:1.∫dx/(a^2 + b^2*x^2)= (1/a^2)∫dx/[1 + (bx/a)^2]= (1/ab)∫d(bx/a)/[1 + (bx/a)^2]= (1/ab)*atan(bx/a) + C
2. w2 =∫dx/(a^2 + b^2*cosx)Set x = 2*atan(u) => dx = 2du/(1 + u^2)cos(x) = 2*cos^2 [(atan(u)] - 1= 2*1/(1 + u^2) - 1= (1 - u^2)/(1 + u^2)
w2 =∫2du/(1 + u^2) / [a + b(1 - u^2)/(1 + u^2)]=∫2du/[a(1 + u^2) - b(1 - u^2)]=∫2du/[(a - b) + (a + b) u^2].....a > b=∫2du/(c + d*u^2).....c = a - b, d = a + b= 2/√(cd) * atan[√(d/c)*u] + C.....by (1)= 2/√(a^2-b^2) * atan{√[(a+b)/(a-b)]tan(x/2)} + C
w3 =∫dx/[(a*cosx)^2 + (b*sinx)^2]=∫dx/[(a*cosx)^2 + b^2(1-cos^2 x)]=∫dx/[b^2 + (a^2 - b^2)cos^2 x]=∫dx/[b^2 - (c*cosx)^2].....c^2 = a^2 - b^2=∫dx/{b^2 - c^2*(1 + cos2x)/2}= 0.5∫d(2x)/{(a^2+b^2)/2 + (c^2/2)*cos(2x)]......y=2x= (1/4)∫dy/(g^2 + h^2*cosy).....g=a^2+b^2, h=a^2-b^2= (1/2√2*b) * atan{(g+h)/(g-h)]tan(y/2)} = (1/2√2*b) * atan{(g+h)/(g-h)]tan(x)}.....x=0~pi= (1/2√2*b) * atan{(g+h)/(g-h)]*(0 - 0)}= (1/2√2*b) * atan{0}= (1/2√2*b) * {0 or pi}= 0 or pi/2√2*b= answers
2015-01-18 15:26:20 補充：
第1題補充:
Total surface of the solid
= Paraboloid + Cylindrical lateral + Botom circle
= (π/6)*[(1 + 4a^2)^1.5 - 1] + 2πa*a + πa^2
= (π/6)*[(1 + 4a^2)^1.5 - 1] + 3πa^2
2015-01-18 16:49:23 補充：
***第一題最後一式中括號裡面多出的-1是甚麼？
Ans: 下限y=0代入
2015-01-18 16:59:36 補充：
***第二題的W2第二段第一行，分子分母同乘以(1 + u^2)變第二行
=∫2du/[a(1 + u^2) - b(1 - u^2)] 這裡中括號中間不是應該＋號嗎？
Ans: 修改
w2 =∫2du/(1 + u^2) / [a^2 + b^2*(1 - u^2)/(1 + u^2)]
=∫2du/[a^2*(1 + u^2) + b^2*(1 - u^2)]
=∫2du/[(a^2 + b^2) + (a^2 - b^2) u^2]
=∫2du/[c^2 + (d*u)^2].....c^2 = a^2 + b^2, d^2 = a^2 - b^2
2015-01-18 17:00:49 補充：
w2= 2/(cd) * atan(d*u/c) + C
= 2/√(a^4-b^4)*atan{√[(a^2-b^2)/(a^2+b^2)]tan(x/2)} + C
2015-01-18 17:04:02 補充：
cos^2 [(atan(u)]＝1/(1 + u^2)怎麼求出？
Ans:
鄰邊=1, 對邊=u => 斜邊=√(1+u^2)
cos(atn(u)=1/√(1+u^2) => cos^2 = 1/(1+u^2)
