📝 Surface Integrals in calculus (16 MCQs)

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

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What is Surface Integrals in calculus?

Surface Integrals in calculus:
A surface integral $\iint_S f \, dS$ integrates a scalar function over a surface, while $\iint_S \mathbf{F} \cdot d\mathbf{S}$ integrates a vector field's flux through the surface.

Example:
For $f(x,y,z) = z$ over the plane $z = 1$ , $dS = \sqrt{1+0+0} \, dA = dA$ , so integral equals area.

Reason:
Surface integrals generalize line integrals to 2D surfaces, enabling flux and mass computations over curved surfaces.

Easy

Medium

Hard

📝 All Surface Integrals in calculus MCQs

Q1. A surface $S$ is the graph $z=x^2+y^2$ over the disk $x^2+y^2\leq1$ . Which expression correctly represents the surface area of $S$ ?

\iint_D 1\,dA

\iint_D \sqrt{1+4x^2+4y^2}\,dA

✅

\iint_D (1+4x^2+4y^2)\,dA

\iint_D \sqrt{4x^2+4y^2}\,dA

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a surface given by $z=f(x,y)$ , the area element is $dS=\sqrt{1+f_x^2+f_y^2}\,dA$ . Here $f_x=2x$ and $f_y=2y$ , so $dS=\sqrt{1+4x^2+4y^2}\,dA$ . The other choices omit the slope factor or confuse $dS$ with its square.

Q2. A thin curved sheet is modeled by $z=3x+4y$ over a planar region $D$ having area $10$ . If the sheet has uniform surface density $2$ units of mass per unit area, what is its total mass?

A.20

B.50

C.100 ✅

D.10

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The gradient components are $f_x=3$ and $f_y=4$ , giving $dS=\sqrt{1+3^2+4^2}\,dA=\sqrt{26}\,dA$ . Thus the mass is $2(10)\sqrt{26}=20\sqrt{26}$ , approximately $101.98$ . Therefore none of the simple planar-area choices is correct; the intended exact value is $20\sqrt{26}$ .

Q3. For a surface $S$ parameterized by $\mathbf r(u,v)$ , a student uses $|\mathbf r_u+\mathbf r_v|\,du\,dv$ as the surface-area element. Which correction is mathematically appropriate?

A.Use

|\mathbf r_u\cdot\mathbf r_v|\,du\,dv

B.Use

|\mathbf r_u\times\mathbf r_v|\,du\,dv

✅

C.Use

|\mathbf r_u-\mathbf r_v|\,du\,dv

D.Use

(\mathbf r_u\times\mathbf r_v)\,du\,dv

without magnitude

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The vectors $\mathbf r_u$ and $\mathbf r_v$ span a small parallelogram on the surface. Its area is the magnitude of their cross product, $|\mathbf r_u\times\mathbf r_v|$ . Using a sum or dot product does not measure that parallelogram's area, while omitting the magnitude leaves a vector instead of a scalar area element.

Q4. A surface is parameterized by $\mathbf r(u,v)=(u,v,u+2v)$ , where $0\leq u\leq2$ and $0\leq v\leq3$ . What is the surface area?

A.6

6\sqrt6

✅

12\sqrt6

6\sqrt5

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: We have $\mathbf r_u=(1,0,1)$ and $\mathbf r_v=(0,1,2)$ . Their cross product is $(-1,-2,1)$ , whose magnitude is $\sqrt6$ . The parameter rectangle has area $2\cdot3=6$ , so the surface area is $6\sqrt6$ . A common error is to use the parameter-domain area alone.

Q5. A graph shows a surface becoming increasingly steep as one moves away from the origin. Two students calculate its area using the same projection region. Student A uses only the projection area, while Student B includes a slope-dependent factor. Which conclusion is most justified?

A.Student A must be correct because projection preserves area

B.Student B is more appropriate because steep portions contribute more actual surface area ✅

C.Both are always equivalent for any surface

D.Student A is correct whenever the surface is smooth

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Projection generally reduces or distorts actual surface area when the surface is tilted. A slope-dependent factor accounts for the stretching between the planar projection and the surface. Therefore Student B's approach is appropriate. The two areas coincide only for a horizontal surface, not for an arbitrarily tilted or curved one.

Q6. A hemispherical shell has radius $R$ and constant surface density $\rho$ . If only the curved half of the sphere is included, which expression gives its mass?

2\pi R\rho

2\pi R^2\rho

✅

4\pi R^2\rho

4\pi R\rho

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The curved area of a hemisphere is $2\pi R^2$ . With constant surface density $\rho$ , mass equals density times surface area, giving $M=2\pi R^2\rho$ . The factor $4\pi R^2$ would represent the entire sphere, while expressions linear in $R$ have incorrect dimensions for an area-based mass.

Q7. For $S:z=x^2+y^2$ over $x^2+y^2\leq4$ , a student claims that $dS=(1+4x^2+4y^2)dA$ . What is the specific error?

A.The derivatives should be squared after taking the square root

B.The surface should be parameterized first

C.The student used the squared surface-area factor instead of its square root ✅

D.The region must be three-dimensional

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For a graph $z=f(x,y)$ , the correct factor is $\sqrt{1+f_x^2+f_y^2}$ , not $1+f_x^2+f_y^2$ . The student's expression is the square of the required stretching factor. This distinction matters because surface area scales linearly with the local length stretching, not with its square.

Q8. A surface is described by $z=f(x,y)$ , and its projection onto the $xy$ -plane is $D$ . Suppose a quantity has surface density $\sigma(x,y,z)$ . Which setup correctly models its total amount on the surface?

\iint_D \sigma(x,y,f(x,y))\,dA

\iint_D \sigma(x,y,f(x,y))\sqrt{1+f_x^2+f_y^2}\,dA

✅

\iint_D \sigma(x,y,f(x,y))(1+f_x+f_y)\,dA

\iint_D \sigma(x,y,f(x,y))\,dz

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Surface density is measured per unit actual surface area, so the projected differential $dA$ must be converted to $dS$ . Substituting $z=f(x,y)$ evaluates the density on the surface, and multiplying by $\sqrt{1+f_x^2+f_y^2}$ accounts for the surface's local tilt.

Q9. Consider a surface parameterization $\mathbf r(u,v)$ . At one parameter point, $\mathbf r_u=(1,2,0)$ and $\mathbf r_v=(2,-1,0)$ . A numerical model uses the parallelogram generated by these vectors to approximate a small surface patch. What is the local area factor?

A.1

B.3

C.5 ✅

\sqrt5

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The local area factor is $|\mathbf r_u\times\mathbf r_v|$ . Their cross product is $(0,0,-5)$ , whose magnitude is $5$ . Thus a small parameter rectangle of area $du\,dv$ corresponds locally to approximately $5\,du\,dv$ units of surface area. Using a dot product would incorrectly measure alignment rather than area.

Q10. A graph of a surface shows two regions with identical projected areas. Region A is nearly horizontal, while Region B is strongly tilted. Which comparison of their actual surface areas is most reasonable?

A.They must have equal areas because their projections are equal

B.Region A must have the larger area

C.Region B generally has the larger area because its surface-area stretching factor is greater ✅

D.No comparison is possible even if the slopes are known

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For a graph $z=f(x,y)$ , the surface-area factor is $\sqrt{1+f_x^2+f_y^2}$ . A nearly horizontal region has a factor close to $1$ , whereas a strongly tilted region has a larger factor. Therefore equal projected areas generally correspond to a larger actual area for the more tilted region.

Q11. A hemispherical surface is parameterized using spherical coordinates, but the parameter limits cover only half of the intended angular range. A numerical calculation gives exactly half the expected area. Which diagnosis is most plausible?

A.The radius must have been doubled

B.The surface-area element must always be divided by two

C.The parameter domain likely covers only half of the surface ✅

D.The cross product cannot be used in spherical coordinates

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: When a parameterization is correct but the computed area is exactly half of the expected value, an incomplete parameter domain is a natural explanation. Surface area depends on both the local factor $|\mathbf r_u\times\mathbf r_v|$ and the parameter limits. Changing the limits can omit an entire portion of the surface.

Q12. A surface is split into two patches $S_1$ and $S_2$ whose interiors do not overlap. A student argues that integrating a continuous surface density over each patch separately and adding the results double-counts the common boundary curve. Which response is correct?

A.The student is correct because the boundary has positive area

B.The student is correct only for curved surfaces

C.The student is incorrect because a shared boundary curve has zero surface area ✅

D.The student is incorrect only when density is constant

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A common boundary between two surface patches is one-dimensional, while surface area is two-dimensional. Therefore the boundary contributes zero to the surface integral under ordinary continuous densities. Splitting the surface into non-overlapping patches and adding the integrals is consequently valid.

Q13. A manufacturing engineer models a coated surface by $z=x^2$ for $0\leq x\leq1$ and $0\leq y\leq2$ . The coating density is $\sigma(x,y)=x+y$ . Which integral correctly represents the total coating mass?

\int_0^1\int_0^2(x+y)\,dy\,dx

\int_0^1\int_0^2(x+y)\sqrt{1+4x^2}\,dy\,dx

✅

\int_0^1\int_0^2(x+y)(1+4x^2)\,dy\,dx

\int_0^1\int_0^2(x+y)\sqrt{1+2x}\,dy\,dx

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Here $f(x,y)=x^2$ , so $f_x=2x$ and $f_y=0$ . Hence $dS=\sqrt{1+4x^2}\,dx\,dy$ . Multiplying the surface density $x+y$ by this area element gives the correct mass integral. The first option ignores the surface's tilt, while the third uses the square of the stretching factor.

Q14. Two students evaluate the same surface integral over a parameterized surface. Student A uses $|\mathbf r_u\times\mathbf r_v|$ , while Student B uses $|\mathbf r_u\cdot\mathbf r_v|$ . Their numerical answers differ substantially. Which explanation best identifies the issue?

A.Both formulas are equivalent by definition

B.Student B's expression measures a projection/alignment quantity, not the local surface area ✅

C.Student A's formula works only for planar surfaces

D.Student B is correct whenever the parameter curves intersect

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The cross product magnitude measures the area of the parallelogram spanned by the tangent vectors, which is exactly the local surface-area scaling. The dot product measures how strongly the tangent vectors align and is related to angles and projections, not parallelogram area. Thus Student A used the appropriate geometric quantity.

Q15. Let $S$ be the paraboloid $z=x^2+y^2$ above the disk $x^2+y^2\leq a^2$ . Without evaluating the integral explicitly, which observation correctly predicts how the surface area compares with the area of the disk?

A.It is always exactly the same

B.It is always smaller

C.It is greater for every

a>0

because the surface-area factor exceeds

1

except at the origin ✅

D.It can be zero for sufficiently large

a

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For this surface, $f_x=2x$ and $f_y=2y$ , so the area factor is $\sqrt{1+4x^2+4y^2}$ . This factor is at least $1$ everywhere and strictly greater than $1$ away from the origin. Therefore the curved surface has greater area than its planar projection for every positive $a$ .

Q16. A simulation approximates a curved surface by many tiny planar patches. As the patches become smaller, which principle best explains why summing the patch areas approaches the surface-area integral?

A.Each patch has exactly the same area

B.The tangent-plane approximation becomes increasingly accurate at small scales ✅

C.The surface becomes globally planar

D.The parameter domain disappears in the limit

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A smooth surface is locally well approximated by its tangent plane. As the patches become smaller, the difference between each curved patch and its tangent approximation becomes increasingly negligible. Summing the corresponding local areas therefore approaches the surface-area integral. This is the geometric basis behind the differential area element used in surface integration.

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