📝 Surface integral definition (16 MCQs)

📖 From Calculus • 16. Topics in vector Calculus • 16 questions available

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What is Surface integral definition?

Surface integral definition:
For a surface $S$ parameterized by $\mathbf{r}(u,v)$ over domain $D$ , $\iint_S f \, dS = \iint_D f(\mathbf{r}(u,v)) \, |\mathbf{r}_u \times \mathbf{r}_v| \, dA$ , where $|\mathbf{r}_u \times \mathbf{r}_v|$ is the surface area element.

Example:
Sphere radius $a$ : $\mathbf{r}(\theta,\phi) = \langle a\sin\phi\cos\theta, a\sin\phi\sin\theta, a\cos\phi \rangle$ , $|\mathbf{r}_\theta \times \mathbf{r}_\phi| = a^2 \sin\phi$ .

Reason:
This definition provides a rigorous framework for integrating over complex surfaces, foundational for flux and applications.

Easy

Medium

Hard

📝 All Surface integral definition MCQs

Q1. A surface $S$ is divided into many small patches, and a scalar function $f(x,y,z)$ is evaluated at a representative point of each patch. Which limiting expression best captures the definition of the surface integral of $f$ over $S$ ?

A.The limit of the sum of

f

values multiplied by the patch areas ✅

B.The limit of the sum of

f

values divided by the patch areas

C.The sum of the patch areas divided by the number of patches

D.The limit of the sum of

f

values multiplied by the patch perimeters

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: A scalar surface integral is defined by accumulating the value of the function over infinitesimal pieces of surface area. Thus each sample value $f(x_k,y_k,z_k)$ is multiplied by the corresponding small area $\Delta S_k$ , and the sum approaches the surface integral as the partition becomes arbitrarily fine. The other choices use inappropriate geometric quantities.

Q2. If $f(x,y,z)=1$ everywhere on a smooth surface $S$ , what does the surface integral $\iint_S f\,dS$ represent?

A.The volume enclosed by

S

B.The surface area of

S

✅

C.The average height of

S

D.The perimeter of the boundary of

S

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: When the integrand is identically $1$ , every small surface patch contributes exactly its area because $1\cdot\Delta S=\Delta S$ . Adding all patches and taking the limiting process therefore gives the total surface area. This distinction is important because a surface integral over $1$ measures area, not enclosed volume or boundary length.

Q3. A material coating has surface density modeled by $\rho(x,y,z)$ kilograms per square meter on a curved sheet $S$ . Which mathematical expression correctly models the total mass?

\iiint_S \rho\,dV

\oint_S \rho\,ds

\iint_S \rho\,dS

✅

\iint_S \rho\,dV

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Surface density is measured per unit surface area, so a small patch with area $\Delta S$ contributes approximately $\rho\,\Delta S$ kilograms. Summing these contributions and taking the limit produces $\iint_S\rho\,dS$ . A volume integral would require a volume density, while a line integral is appropriate for quantities distributed along curves.

Q4. Suppose a surface is partitioned into patches whose areas are $\Delta S_1,\Delta S_2,\ldots,\Delta S_n$ . If the function varies significantly across each patch, which refinement strategy most directly improves the approximation to $\iint_S f\,dS$ ?

A.Increase the patch areas so fewer values are needed

B.Make the patches smaller so the variation of

f

within each patch becomes negligible ✅

C.Replace every patch area with its perimeter

D.Evaluate

f

only at points outside the surface

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The Riemann-sum interpretation of a surface integral depends on approximating the function as nearly constant over each small surface patch. Refining the partition decreases the patch diameters, reducing the error caused by variation of $f$ within each patch. Simply changing the perimeter or using fewer larger patches does not produce the required limiting behavior.

Q5. A surface $S$ consists of two pieces $S_1$ and $S_2$ that meet only along a boundary curve. A scalar function $f$ is defined on both pieces. Which relationship follows directly from the definition of a surface integral?

\iint_S f\,dS=\iint_{S_1}f\,dS-\iint_{S_2}f\,dS

\iint_S f\,dS=\iint_{S_1}f\,dS+\iint_{S_2}f\,dS

✅

\iint_S f\,dS=\iint_{S_1}f\,dS\iint_{S_2}f\,dS

\iint_S f\,dS=0

because the pieces share a boundary

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Surface integration is additive over nonoverlapping surface pieces. The common boundary curve has zero surface area, so it does not create an additional contribution. Therefore, the integral over the complete surface equals the sum of the integrals over its constituent pieces. Subtraction would incorrectly treat the second surface as having an opposite orientation, which is irrelevant for a scalar surface integral.

Q6. A student claims that $\iint_S f\,dS$ must be zero whenever $S$ is a closed surface because every point on the surface has an opposite point. Which response best evaluates the claim?

A.It is always true because opposite points cancel

B.It is false because scalar surface integrals do not generally involve directional cancellation ✅

C.It is true only when

f

is positive

D.It is false only for planar surfaces

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A scalar surface integral accumulates values using positive surface-area elements $dS$ . There is no automatic sign reversal from opposite points or from reversing orientation. Cancellation can occur only because the scalar function itself takes positive and negative values in appropriate amounts. Therefore, closedness or symmetry alone does not force the integral to vanish.

Q7. A thin curved panel occupies a surface $S$ , and its brightness per unit area is represented by $b(x,y,z)$ . A designer wants the total brightness of the panel. Which interpretation of $\iint_S b\,dS$ is most appropriate?

A.It adds brightness over area elements of the panel ✅

B.It computes the volume beneath the panel

C.It measures only the brightness along the panel edge

D.It averages

b

automatically without accounting for area

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The function $b$ represents brightness per unit surface area, so a small patch contributes approximately $b\,\Delta S$ . Summing these contributions across the entire panel gives total brightness. The integral therefore combines both the local intensity and the geometric area of the surface. It is not automatically an average and does not restrict attention to the boundary.

Q8. A student approximates $\iint_S f\,dS$ by selecting points $P_k$ on $S$ and calculating $\sum_{k=1}^n f(P_k)\Delta S_k$ . The student then says the approximation becomes exact simply by increasing $n$ , even if some patches remain large. What is the flaw?

A.The function must always be linear

B.The sample points cannot lie on the surface

C.Increasing the number of patches is insufficient unless the maximum patch size also tends to zero ✅

D.Surface integrals cannot use sums

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The defining limit requires the partition to become fine, meaning the largest patch diameter must approach zero. Merely increasing the number of patches can leave some large patches and therefore prevent the Riemann sums from converging properly. The issue is geometric refinement, not linearity of the function or the admissibility of finite sums.

Q9. A temperature field on a curved metal surface is $T(x,y,z)$ . One region of the surface has high temperature but very small area, while another has moderate temperature over a very large area. Which feature of the surface integral determines which region contributes more strongly?

A.Only the maximum value of

T

B.The product of local temperature and the corresponding surface-area element ✅

C.Only the number of sample points used

D.The orientation of the surface in space

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: A scalar surface integral weights each local function value by the amount of surface area associated with it. Therefore, a very high temperature over a tiny patch may contribute less to the total than a moderate temperature over a large region. This is precisely why the area element $dS$ is essential in the definition.

Q10. Consider a graph of a nonnegative function $f$ drawn over a curved surface $S$ . A shaded portion $S_A$ has twice the area of $S_B$ , while $f$ is approximately the same positive value on both. Which conclusion is most justified?

S_A

contributes approximately twice as much to the surface integral as

S_B

✅

S_B

contributes approximately twice as much because it is smaller

C.Both contributions must be equal regardless of area

D.Neither contribution can be compared without the boundary curve

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For a nonnegative function that is approximately constant with nearly the same value on two regions, the surface integral is approximately the function value multiplied by the region's surface area. Since $S_A$ has twice the area of $S_B$ , its contribution is approximately twice as large. This reasoning follows directly from the Riemann-sum interpretation.

Q11. A graph shows a surface divided into four visible patches. The function values at representative points are approximately $2,4,1,$ and $3$ , while the corresponding patch areas are $5,1,6,$ and $2$ square units. Which Riemann-sum approximation should be used?

2+4+1+3

5+1+6+2

2(5)+4(1)+1(6)+3(2)

✅

2/5+4/1+1/6+3/2

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A surface-integral Riemann sum multiplies each sampled function value by the area of the corresponding surface patch. Thus the approximation is $2(5)+4(1)+1(6)+3(2)$ . Adding only function values ignores patch size, while adding areas ignores the function. Dividing by area also has no role in the basic integral definition.

Q12. A modeler represents the same physical surface using two different partitions: one with many small patches and another with fewer larger patches. Both use representative points and form sums $S_1$ and $S_2$ . What should happen as both partitions are refined appropriately?

A.Both sums should approach the same surface integral ✅

B.The sum with more patches must always approach zero

C.The sum with fewer patches must always be larger

D.The two limits must differ because the partitions are different

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: A well-defined surface integral is independent of the particular sequence of increasingly fine partitions, provided the function is suitably integrable and the maximum patch size tends to zero. Different partitions may give different finite approximations, but under the limiting process they converge to the same value. This partition independence is fundamental to the Riemann-sum definition.

Q13. A student reasons: 'Because $dS$ represents an infinitesimal surface area, I can replace it by $dx\,dy\,dz$ when integrating over a surface.' Which diagnosis is most accurate?

A.Correct, because all infinitesimal quantities are equivalent

B.Incorrect, because

dx\,dy\,dz

represents a volume element rather than a surface-area element ✅

C.Correct only for closed surfaces

D.Incorrect only when

f

is negative

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The distinction between geometric dimensions is essential. The element $dS$ measures infinitesimal two-dimensional area on a surface, whereas $dV=dx\,dy\,dz$ measures three-dimensional volume. Replacing $dS$ by $dV$ changes the type of integral and the physical dimensions of the result, so the substitution is generally invalid.

Q14. Suppose $f$ is nonnegative on $S$ , and a second function $g$ satisfies $g(x,y,z)\ge f(x,y,z)$ at every point of $S$ . Without evaluating either integral, what can be concluded?

\iint_S g\,dS\le\iint_S f\,dS

\iint_S g\,dS=\iint_S f\,dS

always

\iint_S g\,dS\ge\iint_S f\,dS

✅

D.No comparison is possible because

dS

is variable

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Surface integration preserves pointwise inequalities because the area element $dS$ is nonnegative. Since $g\ge f$ at every point, each local contribution $g\,dS$ is at least as large as $f\,dS$ . Summing over all patches and passing to the limit therefore gives $\iint_Sg\,dS\ge\iint_Sf\,dS$ .

Q15. A sensor is mounted on a surface $S$ . Its response per unit area is modeled by $f$ , but calibration shows the response is multiplied everywhere by a constant factor $c$ . How should the total modeled response change?

A.It becomes

c\iint_S f\,dS

✅

B.It becomes

\iint_S f\,dS+c

C.It becomes

\frac{1}{c}\iint_S f\,dS

D.It becomes

\iint_S f\,dS+c\,dS

without integration

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A constant multiplier can be factored through the surface integral. If the local response changes from $f$ to $cf$ , then every Riemann-sum contribution changes from $f(P_k)\Delta S_k$ to $c f(P_k)\Delta S_k$ . Taking the limit gives $c\iint_Sf\,dS$ , provided the integral exists.

Q16. Let $S$ be a surface of fixed area $A$ , and suppose $f$ is continuous and satisfies $0\le f\le M$ everywhere on $S$ . Which bound follows directly from the definition of the surface integral?

0\le\iint_S f\,dS\le MA

✅

M\le\iint_S f\,dS\le A

0\le\iint_S f\,dS\le M+A

A\le\iint_S f\,dS\le MA^2

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Because $0\le f\le M$ , every small contribution satisfies $0\le f\,\Delta S\le M\Delta S$ . Adding over all patches gives $0\le\sum f_k\Delta S_k\le M\sum\Delta S_k$ . In the limit, the total surface area is $A$ , producing $0\le\iint_Sf\,dS\le MA$ . This bound also provides a useful consistency check for models.

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