Compute the surface integral of the scalar field U(x,y) = (x-x’y-y) over the surface S represented by a square with x ∈ [0,1] and y ∈ [0,1] located on the plane z = 0.
Compute the surface integral of the scalar field U(x,y) = (x-x’y-y) over the surface S represented by a square with x ∈ [0,1] and y ∈ [0,1] located on the plane z = 0.
The Correct Answer and Explanation is:
We are given a scalar field: U(x,y)=x−x′y−yU(x, y) = x – x’y – y
and a surface SS defined as the unit square in the xyxy-plane, i.e., S:{(x,y,z)∣x∈[0,1],y∈[0,1],z=0}S: \{(x, y, z) \mid x \in [0, 1], y \in [0, 1], z = 0\}
The problem is to compute the surface integral of U(x,y)U(x, y) over this surface: ∬SU(x,y) dS\iint_S U(x, y)\, dS
Step 1: Interpret the scalar field
First, clarify the scalar field U(x,y)=x−x′y−yU(x, y) = x – x’y – y. The variable x′x’ is unclear; in mathematical notation, this often denotes a derivative or another variable. If this is a typo and x′x’ actually means just xx, then: U(x,y)=x−xy−yU(x, y) = x – xy – y
Assuming this interpretation, the function simplifies to: U(x,y)=x(1−y)−yU(x, y) = x(1 – y) – y
Step 2: Set up the surface integral
Since the surface lies on the plane z=0z = 0, and dS=dx dydS = dx\,dy, the surface integral becomes a double integral over the unit square: ∬SU(x,y) dx dy=∫01∫01(x(1−y)−y) dx dy\iint_S U(x, y)\, dx\,dy = \int_0^1 \int_0^1 \left(x(1 – y) – y\right)\, dx\,dy
Step 3: Compute the integral
∫01∫01(x(1−y)−y) dx dy=∫01[∫01x(1−y) dx−∫01y dx]dy\int_0^1 \int_0^1 \left(x(1 – y) – y\right)\, dx\,dy = \int_0^1 \left[ \int_0^1 x(1 – y) \, dx – \int_0^1 y \, dx \right] dy
Evaluate the inner integrals:
- ∫01x(1−y) dx=(1−y)∫01x dx=(1−y)⋅12=1−y2\int_0^1 x(1 – y)\, dx = (1 – y) \int_0^1 x\, dx = (1 – y)\cdot\frac{1}{2} = \frac{1 – y}{2}
- ∫01y dx=y⋅(1−0)=y\int_0^1 y\, dx = y \cdot (1 – 0) = y
Now evaluate the outer integral: ∫01[1−y2−y]dy=∫01(1−y−2y2)dy=∫01(1−3y2)dy=12∫01(1−3y) dy\int_0^1 \left[ \frac{1 – y}{2} – y \right] dy = \int_0^1 \left( \frac{1 – y – 2y}{2} \right) dy = \int_0^1 \left( \frac{1 – 3y}{2} \right) dy = \frac{1}{2} \int_0^1 (1 – 3y)\, dy =12[y−32y2]01=12(1−32)=12⋅(−12)=−14= \frac{1}{2} \left[ y – \frac{3}{2}y^2 \right]_0^1 = \frac{1}{2} \left( 1 – \frac{3}{2} \right) = \frac{1}{2} \cdot \left( -\frac{1}{2} \right) = -\frac{1}{4}
✅ Final Answer:
−14\boxed{-\frac{1}{4}}
Explanation
The given problem requires evaluating a surface integral of a scalar field over a flat region. Surface integrals of scalar fields represent the accumulation of a quantity (like temperature or density) over a surface area. In this case, the scalar field U(x,y)=x−x′y−yU(x, y) = x – x’y – y is defined over the unit square in the xyxy-plane, where x,y∈[0,1]x, y \in [0, 1].
A crucial step was interpreting the notation. The term x′x’ is ambiguous—often it denotes a derivative, but in this context, we assumed it’s a typographical error and interpreted it simply as xx. Thus, we rewrote the field as U(x,y)=x(1−y)−yU(x, y) = x(1 – y) – y, a more standard scalar expression.
Since the surface lies flat on the z=0z = 0 plane and spans the square [0,1]×[0,1][0,1] \times [0,1], the surface integral reduces to a standard double integral over that square. That is, the surface element dSdS becomes dx dydx\,dy, and we integrate the function directly over the 2D region.
The integral was computed step-by-step: first integrating with respect to xx, then with respect to yy. The result after simplification was: ∬SU(x,y) dx dy=−14\iint_S U(x, y)\, dx\,dy = -\frac{1}{4}
This value represents the net accumulation of the scalar field over the unit square. The negative result indicates that, on average, the field has a net “negative contribution” over the square—likely because the negative terms dominate over most of the domain.
This type of integral is fundamental in physics and engineering, where scalar fields describe quantities distributed over surfaces, such as pressure on a plate or temperature across a wall.
