Calculate the iterated integral.?

Question

1. S (integral), pie 3.14, I didn't have a pie sign.
The R should be under the SS

S 0 to 2 S 0 to pie/2 x sin y dy dx


2. S 0 to 2 S 0 to 1 (2x + y)^ 8 dx dy


3. SS (6x^2y^3 - 5y^4) dA, R= {(x,y)| 0 <= x <= 3, 0 <=y<=1}
R

4. SS xye^x^2y dA, R= [0,1] x [0,2]
R


5. Find the volume of the solid that lies under the plane 3x + 1y + z= 12 and above the rectangle R= {(x,y) | 0<= x <= 1, -2 <= y<=3}

Answer 1

1. ∫0 to 2 ∫ 0 to pi/2 (x siny dy dx)
= (∫0 to 2 x dx) * (∫0 to pi/2 siny dy) = 2 * 1 = 2

2.
∫0 to 2 ∫ 0 to 1 (2x + y)^8 dx dy
= ∫0 to 2 [∫0 to 1 (2x + y)^8dy] dx
= ∫0 to 2 [1/9 ((2x + 1)^9 - (2x + 0)^9] dx
= 1/9 ∫0 to 2 [(2x + 1)^9 - (2x)^9] dx
= (1/9) * (1/10) * (1/2) [(2*2+1)^10 - (2*2)^10 - (2*0+1)^10 + (2*0)^10]
= 1/180 (5^10 - 4^10 - 1^10 + 0)
= 8717048/180 = 2179262/45

3.
∫0 to 1 [∫0 to 3 (6x²y³ - 5y^4)dx]dy
= ∫ 0 to 1 (2*3³y³ - 3*5y^4 - 0) dy
= ∫0 to 1 (54y³ - 15y^4)dy
= 27/2 * 1^4 - 3*1^5 = 27/2 - 3 = 21/2

4.
(∫0 to 1 xe^xdx) * (∫0 to 2 ??? dy)
the question isn't clear !!?


5. ∫0 to 1[∫-2 to 3 (∫0 to (12-3x-y) dz) dy ] dx
= ∫0 to 1 [∫-2 to 3 (12-3x-y)dy ]dx
= ∫0 to 1 [(12*3 - 3x*3 - 3²/2) - (12*-2 - 3x*-2 - (-2)²/2]
=∫0 to 1 (26 - 9x - 9/2 + 24 - 6x + 2)
= ∫0 to 1 (-15x + 95/2)
= -15/2 + 95/2 = 40

Answer 2

i've got chop up them up, i wish it particularly is clever, any questions, ask ?_0^2 dy = _0^2 [y] = 2 - 0 = 2 ?_0^y ((y^2) * (e^xy)) dx = _0^y [(y^2)*((e^xy)/y)] = (y^2)*((e^(y^2))/(y)) - one million = y*(e^(y^2)) - one million Multiply the two jointly 2*(y*(e^(y^2)) - one million) = (2y*e^(y^2)) - 2 kb and that i are answering distinctive questions. I misinterpret your query, as an hassle-free critical distinctive, use his.