Find the double integral of 2xdydz + ydzdx + zdxdy, where the curve equation is Z = x ^ 2 + y ^ 2 (0

Find the double integral of 2xdydz + ydzdx + zdxdy, where the curve equation is Z = x ^ 2 + y ^ 2 (0

This is not a double integral, but a surface integral of the second kind, using Gauss formula
Complement plane, z = 1, x ^ 2 + y ^ 2 ≤ 1, take the upper side
In this way, the two surfaces are merged into a closed surface
∫∫ 2xdydz+ydzdx+zdxdy
=∫∫∫ (2+1+1)dxdydz
=4∫∫∫1dxdydz
Let's use cylindrical coordinates
=4∫∫∫rdzdrdθ
=4∫[0→2π]dθ∫[0→1]rdr∫[r²→1]dz
=8π∫[0→1]r(1-r²)dr
=8π[(1/2)r^2-(1/4)r^4] |[0→1]
=2π
Now we calculate the integral on the plane of the complement
∫∫ 2xdydz+ydzdx+zdxdy
=∫∫ 1 DXDY integral region is: x ^ 2 + y ^ 2 ≤ 1
=π
So the result of this problem is: the original formula = 2 π - π = π

The double integral x ^ 2 dydz + y ^ 2 dzdx + Z ^ 2 DXDY on S, where s is the outside of the cylindrical surface x ^ 2 + y ^ 2 = a ^ 2 (0 ≤ Z ≤ h)

First of all, I will tell you that this problem is not a double integral, but a second kind of surface integral. We need to use Gauss formula, but Gauss formula requires that the integral surface be closed

∫ (x ^ 2-yz) dydz + (y ^ 2-zx) dzdx + 2zdxdy, where the integral region is Z = 1 - √ (x ^ 2 + y ^ 2) where (z > = 0)

Find the surface integral ∫ ∫ (x ^ 2 * y ^ 2 * z) DXDY of a pair of coordinates, where s is the lower side of the lower part of the spherical surface x ^ 2 + y ^ 2 + Z ^ 2 = R ^ 2

This is the surface integral of the second type. The display expression of the surface is Z = - radical (R ^ 2-x ^ 2-y ^ 2)
The third component of the normal vector is - 1. Let d be x ^ 2 + y ^ 2

Let d be the region bounded by the lines y = 1, 2x-y + 3 = 0 and 2x-y-3 = 0 on the xoy plane, and find ∫ (2x-y) DXDY
In the D domain
The title is the calculation of double integral in Higher Mathematics: ∫ (2x-y) DXDY, D is the area bounded by y = 1, 2x-y + 3 = 0, x + Y-3 = 0

First product y,
∫∫ (2x-y) dxdy
=∫[0→1] dx∫[3-x→2x+3] (2x-y) dy
=∫[0→1] [2xy-(1/2)y²] |[3-x→2x+3]dx
=∫[0→1] [2x(2x+3)-(1/2)(2x+3)²-2x(3-x)+(1/2)(3-x)²] dx
=∫[0→1] [4x²+6x-(1/2)(2x+3)²-6x+2x²+(1/2)(3-x)²] dx
=[(4/3)x³+3x²-(1/12)(2x+3)³-3x²+(2/3)x³-(1/6)(3-x)³] |[0→1]
=4/3 + 3 - 125/12 - 3 + 2/3 - 4/3 + 27/12 + 27/6
=-3
If you don't understand, please ask. If you can solve the problem, please click "select as satisfactory answer" below

If f (x, y) is continuous in the bounded domain D, f (x, y) = 2x + ∫ f (x, y) DXDY, then the partial derivative of F with respect to X is

Note: the result of double integral is a number, which is a constant, so
F(x,y)=2x+C
Fx(x,y)=2

∫∫ (Y / x) DXDY, where D is a planar region bounded by y = x, y = x ^ 3
Calculate double integral

∫∫(y/x)dxdy
=∫dx∫(y/x)dy+∫dx∫(y/x)dy
=∫1/2(x-x^5)dx+∫1/2(x^5-x)dx
=1/2(x²/2-x^6/6)|+1/2(x^6/6-x²/2)|
=1/6+1/6
=1/3

∫∫ Ye ^ (XY) DXDY, where D is bounded by the curves xy = 1 and x = 1, x = 2, and y = 2
The first step is to elaborate on the y-integral

The original formula = ∫ [1,2] DX ∫ [1 / x, 2] Ye ^ (XY) dy = ∫ [1,2] DX ∫ [1 / x, 2] y / Xe ^ (XY) d (XY) in the first integral of Y, X is a constant = ∫ [1,2] 1 / xdx ∫ [1 / x, 2] yde ^ (XY) 1 / x, which can be regarded as a constant mentioned before the integral sign (in the integral of DX) = ∫ [1,2] 1 / xdx (Ye ^ (XY) | [1 / x, 2] -∫ [1

∫∫ e ^ (Y-X / y + x) DXDY, where D is a closed region bounded by x-axis, Y-axis and straight line x + y = 2

(- 1) & # 178; + (1 / 2) negative quadratic minus root 3 / 3

=1 + {1/[(½)²]} - 3/√3
=1 + 4 - √3
=5 - √3
For you to understand the order, add a few more brackets
I don't know if I understand the last item right. If it's √ (3 / 3), it's better:
=1 + {1/[(½)²]} - √(3/3)
=1 + 4 - √1
=5 - 1
=4