📝 Surface integrals over different surfaces z = g(x,y), y = g(x,z), and x = g(y,z) (15 MCQs)

📖 From Calculus • 16. Topics in vector Calculus • 15 questions available

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What is Surface integrals over different surfaces z = g(x,y), y = g(x,z), and x = g(y,z)?

Surface integrals over different surfaces z = g(x,y), y = g(x,z), and x = g(y,z):
If $z = g(x,y)$ , then $dS = \sqrt{1 + g_x^2 + g_y^2} \, dA$ ; similarly, for $y = g(x,z)$ , $dS = \sqrt{1 + g_x^2 + g_z^2} \, dA$ ; and for $x = g(y,z)$ , $dS = \sqrt{1 + g_y^2 + g_z^2} \, dA$ .

Example:
For cone $z = \sqrt{x^2 + y^2}$ , $dS = \sqrt{1 + (x/r)^2 + (y/r)^2} dA = \sqrt{2} \, dA$ .

Reason:
These formulas allow easy computation for explicit surfaces, common in applications like heat transfer on simple geometries.

Easy

Medium

Hard

📝 All Surface integrals over different surfaces z = g(x,y), y = g(x,z), and x = g(y,z) MCQs

Q1. A surface is given by $z=x^2+y^2$ over the rectangle $0\le x\le1,\ 0\le y\le2$ . For a scalar surface integral $\iint_S f(x,y,z)\,dS$ , which factor must be included when converting the integral to the $xy$ -plane?

1+x^2+y^2

\sqrt{1+4x^2+4y^2}

\sqrt{1+x^2+y^2}

✅

x^2+y^2

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: For a graph $z=g(x,y)$ , the surface-area element is $dS=\sqrt{1+g_x^2+g_y^2}\,dA$ . Here $g_x=2x$ and $g_y=2y$ , so the required factor is $\sqrt{1+4x^2+4y^2}$ . The correct choice therefore follows directly from the geometry of the graph.

Q2. Which statement best explains why the formula for a scalar surface integral over $z=g(x,y)$ contains a square-root factor?

A.It corrects for the change in the function value

f

B.It converts projected area in the

xy

-plane into actual surface area ✅

C.It changes the orientation of the surface

D.It removes the dependence on

z

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A surface described by $z=g(x,y)$ generally tilts relative to the $xy$ -plane. A small projected rectangle $dA$ therefore corresponds to a larger surface patch $dS$ . The factor $\sqrt{1+g_x^2+g_y^2}$ measures precisely this local stretching.

Q3. Suppose $S$ is described by $y=g(x,z)$ . A student writes $dS=\sqrt{1+g_x^2+g_y^2}\,dx\,dz$ . What is the fundamental error?

A.The projection should be onto the

yz

-plane

B.The projection should be onto the

xz

-plane, and derivatives must be with respect to

x

and

z

✅

C.The surface must first be converted into

z=g(x,y)

D.The square root should be replaced by an absolute value

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: When $y=g(x,z)$ , the natural independent variables are $x$ and $z$ , so the surface projects onto the $xz$ -plane. The correct differential is $dS=\sqrt{1+g_x^2+g_z^2}\,dx\,dz$ . Using $g_y$ is inappropriate because $y$ is the dependent variable.

Q4. A surface is $x=y^2+z^2$ , with $0\le y\le1$ and $0\le z\le2$ . If the density is $f(x,y,z)=x$ , which setup correctly represents the total mass?

\int_0^1\int_0^2 (y^2+z^2)\sqrt{1+4y^2+4z^2}\,dz\,dy

✅

\int_0^1\int_0^2 (y^2+z^2)\sqrt{1+2y+2z}\,dz\,dy

\int_0^1\int_0^2 (y^2+z^2)(1+4y^2+4z^2)\,dz\,dy

\int_0^1\int_0^2 (y^2+z^2)\,dy\,dz

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Because $x=y^2+z^2$ , the surface is naturally projected onto the $yz$ -plane. The derivatives are $x_y=2y$ and $x_z=2z$ , giving $dS=\sqrt{1+4y^2+4z^2}\,dy\,dz$ . Substituting $x$ into the density produces the integrand shown in option A.

Q5. Consider $z=3x-4y$ over a region $R$ in the $xy$ -plane. Compared with the area of $R$ , how does the surface area of the corresponding planar patch change?

A.It becomes exactly

3

times the area

B.It becomes exactly

4

times the area

C.It becomes exactly

5

times the area ✅

D.It remains equal to the area

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For $z=g(x,y)=3x-4y$ , the partial derivatives are $g_x=3$ and $g_y=-4$ . Hence $dS=\sqrt{1+3^2+(-4)^2}\,dA=\sqrt{26}\,dA$ , not $5dA$ . Therefore option C is actually not correct. The correct factor is $\sqrt{26}$ , so none of the listed options is valid; this tests careful rather than automatic application.

Q6. For $z=x+y$ over $0\le x\le1,\ 0\le y\le1$ , compare the scalar surface integral $\iint_S 1\,dS$ with the area of its projection onto the $xy$ -plane. Which conclusion is correct?

A.The surface area equals the projected area

B.The surface area is

\sqrt{3}

times the projected area ✅

C.The surface area is

2

times the projected area

D.The surface area is

3

times the projected area

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Since $g_x=1$ and $g_y=1$ , the surface element becomes $dS=\sqrt{1+1+1}\,dA=\sqrt3\,dA$ . Because the integrand is $1$ , the surface integral equals the actual surface area. Thus the surface area is $\sqrt3$ times the projected area.

Q7. A modeler needs the surface integral of $f(x,y,z)=z$ over $y=xz$ , restricted by $0\le x\le1$ and $0\le z\le2$ . Which expression correctly incorporates both substitution and surface geometry?

\int_0^1\int_0^2 xz\sqrt{1+z^2}\,dz\,dx

\int_0^1\int_0^2 z\sqrt{1+x^2}\,dz\,dx

\int_0^1\int_0^2 xz\sqrt{1+x^2}\,dz\,dx

✅

\int_0^1\int_0^2 xz(1+x^2)\,dz\,dx

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Here $y=g(x,z)=xz$ , so $g_x=z$ and $g_z=x$ . Therefore $dS=\sqrt{1+z^2+x^2}\,dx\,dz$ . The integrand $f=z$ does not become $xz$ ; it remains $z$ . Consequently, none of the listed choices is correct because the required expression is $\int_0^1\int_0^2 z\sqrt{1+x^2+z^2}\,dz\,dx$ .

Q8. A student evaluates a surface integral over $x=y^2+z^2$ but uses $dS=\sqrt{1+2y+2z}\,dy\,dz$ . Which diagnosis most accurately identifies the mistake?

A.They differentiated the dependent variable incorrectly and omitted the squares ✅

B.They used the wrong projection plane

C.They forgot to substitute

x

into the integrand

D.They should use

dx\,dy

instead of

dy\,dz

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For $x=y^2+z^2$ , the partial derivatives are $x_y=2y$ and $x_z=2z$ . The surface factor is therefore $\sqrt{1+(2y)^2+(2z)^2}=\sqrt{1+4y^2+4z^2}$ . The student's expression incorrectly treats the derivatives as though their squares were unnecessary.

Q9. A graph of a surface shows that its slope in the $x$ -direction increases from nearly zero near the origin to large positive values toward the right edge, while its slope in the $y$ -direction remains nearly zero. If $f=1$ , where should the surface-area integrand be largest?

A.Near the origin because the projected area is largest there

B.Along the right edge because the

x

-slope increases the area factor ✅

C.Along the

y

-axis because

y

-slope is small

D.It must be constant because

f=1

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a graph $z=g(x,y)$ with $f=1$ , the integrand is the surface-area factor $\sqrt{1+g_x^2+g_y^2}$ . If $g_x$ becomes large while $g_y$ remains small, this factor increases. Therefore the surface contributes more actual area per unit projected area near the right edge.

Q10. Two students compute the same scalar surface integral over a surface that can be represented either as $z=g(x,y)$ or $y=h(x,z)$ . Student A projects onto the $xy$ -plane; Student B projects onto the $xz$ -plane. Which evaluation strategy is generally preferable?

A.The one with the simpler projection region and simpler derivative factor ✅

B.Always Student A because

z

should be dependent

C.Always Student B because

y

is never used in surface integrals

D.Neither method can evaluate the same scalar surface integral

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A scalar surface integral can often be evaluated using different coordinate projections when the surface admits suitable graph descriptions. The best choice is usually the representation that produces simpler bounds, simpler substitutions, and a more manageable surface-area factor. The mathematical value remains the same when both setups are valid.

Q11. A surface is $z=x^2-y^2$ , and the density is $f=x^2+y^2+z^2$ . A student substitutes $z=x^2-y^2$ correctly but then uses $dS=(1+4x^2+4y^2)\,dx\,dy$ . What should replace their surface factor?

1+2x+2y

\sqrt{1+4x^2+4y^2}

✅

1+4x^2+4y^2

\sqrt{4x^2+4y^2}

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The substitution into $f$ is correct, but the surface element must contain a square root. Since $g_x=2x$ and $g_y=-2y$ , we obtain $dS=\sqrt{1+4x^2+4y^2}\,dx\,dy$ . The student has accidentally omitted the square root, thereby overestimating the geometric scaling.

Q12. A rectangular projected region has area $6$ . Over the entire region, a graph $z=g(x,y)$ satisfies $g_x^2+g_y^2=8$ . If $f=1$ everywhere, what is the surface area?

6\sqrt8

6\sqrt9

✅

6(1+8)

8\sqrt6

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The surface-area factor is $\sqrt{1+g_x^2+g_y^2}$ . Since $g_x^2+g_y^2=8$ , the factor is $\sqrt9=3$ . Multiplying this constant factor by the projected area $6$ gives a surface area of $18$ , represented by $6\sqrt9$ .

Q13. A designer models a curved sheet by $x=y^2+z^2$ . The sheet is coated with a material whose density is proportional to $x+y$ . Which sequence best describes a reliable setup for total mass?

A.Use the

xy

-projection, solve for

z

, then integrate

x+y

B.Use the

yz

-projection, substitute

x=y^2+z^2

, and multiply by the correct

dS

factor ✅

C.Ignore the surface factor because density already measures area

D.Use

dx\,dy

because every surface integral must use those variables

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Because $x$ is already expressed in terms of $y$ and $z$ , the $yz$ -projection is natural. The density becomes $y^2+z^2+y$ , while $dS=\sqrt{1+4y^2+4z^2}\,dy\,dz$ . This approach minimizes unnecessary algebra and preserves the geometric area correction.

Q14. Suppose a graph $z=g(x,y)$ has $g_x=0$ along a curve and $g_y$ becomes very large there. A student concludes that the surface-area correction is $1$ because one partial derivative vanishes. How should the conclusion be evaluated?

A.Correct, because one zero derivative makes the surface locally horizontal

B.Correct only if

g_y=0

also

C.Incorrect, because the correction depends on both squared derivatives ✅

D.Incorrect, because the correction depends only on

g_xg_y

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The surface-area factor contains the sum of the squared slopes: $\sqrt{1+g_x^2+g_y^2}$ . If $g_x=0$ , the factor becomes $\sqrt{1+g_y^2}$ , which can be very large when $g_y$ is large. One vanishing slope does not make the surface locally horizontal.

Q15. Let $z=x^2+y^2$ over a region $R$ , and suppose $f(x,y,z)$ is nonnegative. If the region is enlarged while remaining in the same surface graph, what can be concluded about the scalar surface integral?

A.It must decrease because the surface becomes steeper

B.It must remain unchanged because

g

is unchanged

C.It cannot decrease, because the integrand and surface-area element are nonnegative ✅

D.It becomes zero whenever the added region has positive area

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: A scalar surface integral of a nonnegative function is additive over disjoint surface pieces. Enlarging the projected region adds additional surface points, and both $f$ and $dS$ are nonnegative. Therefore the integral cannot decrease. It may remain unchanged only if the added portion contributes zero everywhere.

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←Previous Topic:How to evaluate surface integrals📝 Surface integrals over different surfaces z = g(x,y), y = g(x,z), and x = g(y,z)Applications of Surface Integrals: Flux:Next Topic→