Multivariable Calculus Quiz: Using Symmetry

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Using SymmetryQuestion 1 of 7

Consider the double integral $\iint_D f(x,y) \, dA$ where $f(x,y) = x^3\sin(y^2) + y^2\cos(x^4) + xy\sin(x^2 + y^2)$ and $D$ is the disk $x^2 + y^2 \leq 16$ . A student claims the integral equals zero by symmetry arguments. Which part of the student's reasoning, if any, is incorrect?

The reasoning is correct; all three terms integrate to zero due to odd symmetry properties over the circular disk

Only the first term

x^3\sin(y^2)

integrates to zero; the other terms contribute non-zero values

The third term

xy\sin(x^2 + y^2)

does not integrate to zero because

\sin(x^2 + y^2)

breaks the required symmetry

The second term

y^2\cos(x^4)

cannot be zero because

y^2

is always non-negative

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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Using Symmetry

Practice Using Symmetry in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Using Symmetry, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The reasoning is correct; all three terms integrate to zero due to odd symmetry properties over the circular disk
Only the first term $x^3\sin(y^2)$ integrates to zero; the other terms contribute non-zero values (correct answer)
The third term $xy\sin(x^2 + y^2)$ does not integrate to zero because $\sin(x^2 + y^2)$ breaks the required symmetry
The second term $y^2\cos(x^4)$ cannot be zero because $y^2$ is always non-negative

Explanation: The first term x³sin(y²) is odd in x, so it integrates to zero over the disk which is symmetric about the y-axis. The third term xy·sin(x²+y²) is odd in both x and y individually (since sin(x²+y²) is even in both variables), so it also integrates to zero due to the symmetry of the circular disk. However, the second term y²cos(x⁴) is even in y, and while cos(x⁴) is even in x, the product y²cos(x⁴) doesn't have the right symmetry to make its integral zero over the disk.

Question 2

The triple integral $\iiint_E \frac{xyz}{(x^2 + y^2 + z^2)^{3/2}} \, dV$ is evaluated over the solid ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \leq 1$ where $a, b, c > 0$ . Using symmetry considerations, what is the most direct way to determine this integral's value?

Convert to elliptical coordinates and recognize that the odd symmetry in each variable makes the integral zero (correct answer)
The integral is zero only if $a = b = c$ ; otherwise it must be computed directly using elliptical coordinates
Split the integral into eight octants and use the fact that contributions from opposite octants cancel exactly
The singularity at the origin prevents the use of symmetry arguments, requiring careful limit analysis instead

Explanation: The integrand xyz/(x²+y²+z²)^(3/2) is odd in each variable x, y, and z separately. The ellipsoid is symmetric about each coordinate plane (the yz-plane, xz-plane, and xy-plane). When integrating an odd function over a symmetric domain, the integral equals zero regardless of the values of a, b, c. The fact that the ellipsoid isn't spherical doesn't matter - the symmetry argument still applies. The integrand does have a singularity at the origin, but this doesn't prevent the symmetry argument since the singularity has measure zero.

Question 3

The line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ where $\mathbf{F}(x,y) = (x^3 - 3xy^2, y^3 - 3x^2y)$ is computed along the closed curve $C$ that forms the boundary of the region $\{(x,y): 1 \leq x^2 + y^2 \leq 4\}$ , with the outer circle traversed counterclockwise and inner circle clockwise. A student uses Green's theorem and claims the answer is zero due to symmetry. What is the flaw in this reasoning?

Green's theorem cannot be applied because the region is not simply connected due to the inner boundary
The vector field $\mathbf{F}$ is conservative, making the integral zero, but this isn't due to the symmetry of the region (correct answer)
The curl of $\mathbf{F}$ is not zero everywhere, so Green's theorem gives a non-zero result despite the radial symmetry
The symmetry argument fails because the orientations of the inner and outer boundaries have opposite signs

Explanation: Computing the curl: ∂(y³-3x²y)/∂x - ∂(x³-3xy²)/∂y = -6xy - (-6xy) = 0. Since curl F = 0, the vector field is conservative and the line integral around any closed curve is zero. However, this is not due to symmetry of the region but due to the conservative nature of the field. Green's theorem can be applied to the annular region (it's simply connected when properly oriented). The student's conclusion is correct but the reasoning about symmetry is wrong - it's the zero curl that makes the integral zero.

Question 4

Consider the triple integral $\iiint_E (x^4 + y^4 + z^4 + 2x^2y^2 + 2y^2z^2 + 2x^2z^2) \, dV$ over the solid sphere $x^2 + y^2 + z^2 \leq R^2$ . Using symmetry arguments, which expression correctly represents the simplified form of this integral?

$3\iiint_E x^4 \, dV + 6\iiint_E x^2y^2 \, dV$ by rotational symmetry of the sphere
$\frac{1}{3}\iiint_E (x^2 + y^2 + z^2)^2 \, dV$ since the integrand equals the square of the distance function
$3\iiint_E x^4 \, dV + 2\iiint_E x^2y^2 \, dV$ because symmetry makes the cross terms equal
$\iiint_E (x^2 + y^2 + z^2)^2 \, dV$ because the integrand can be factored as the square of the radius (correct answer)

Explanation: The integrand x⁴ + y⁴ + z⁴ + 2x²y² + 2y²z² + 2x²z² is exactly (x² + y² + z²)². This can be verified by expanding (x² + y² + z²)² = x⁴ + y⁴ + z⁴ + 2x²y² + 2x²z² + 2y²z². So the integral becomes ∭ (x² + y² + z²)² dV = ∭ r⁴ dV in spherical coordinates. Option A incorrectly tries to use symmetry to reduce terms. Option B has the wrong coefficient. Option C makes an error in counting the cross terms after applying symmetry.

Question 5

A flux integral $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS$ is computed where $\mathbf{F}(x,y,z) = (x^3 + yz, y^3 + xz, z^3 + xy)$ and $S$ is the surface of the cube $[0,2] \times [0,2] \times [0,2]$ with outward normal. A student attempts to use symmetry by claiming the cube is symmetric about the planes $x = 1, y = 1, z = 1$ . Which statement best describes the validity of applying symmetry arguments here?

Symmetry arguments apply directly since the cube and vector field are both symmetric about the planes $x = 1, y = 1, z = 1$
Symmetry arguments partially apply; some terms can be simplified but the analysis is more complex than standard symmetry cases (correct answer)
Symmetry arguments don't apply because the cube $[0,2]^3$ is not centered at the origin like standard symmetry scenarios
The divergence theorem should be used instead, where $\nabla \cdot \mathbf{F} = 3x^2 + 3y^2 + 3z^2$ simplifies the calculation

Explanation: While the cube [0,2]³ is symmetric about the planes x=1, y=1, z=1, applying symmetry arguments to eliminate flux contributions is complex. The vector field components like x³ behave differently on opposite faces (x=0 vs x=2) when combined with the outward normal directions. The mixed terms yz, xz, xy add further complexity. Although symmetry can provide some insights, the standard symmetry elimination techniques don't apply straightforwardly here, making direct computation or divergence theorem more practical approaches.

Question 6

Consider the line integral $\oint_C (x^2y - y^3) dx + (xy^2 + x^3) dy$ where $C$ is the boundary of the region $R = \{(x,y): x^2 + y^2 \leq 4, y \geq 0\}$ traversed counterclockwise. Using symmetry properties of the integrand and region, which approach most efficiently determines the integral's value?

Apply Green's theorem directly since the partial derivatives create a simple integrand over the semicircular region (correct answer)
Use symmetry to show the integral equals twice the integral over the right half of the semicircle only
Recognize that odd-even symmetry properties make the integral zero without further calculation
Split into semicircular arc and diameter segments, using symmetry to eliminate the arc contribution

Explanation: Green's theorem gives ∮_C P dx + Q dy = ∬_R (∂Q/∂x - ∂P/∂y) dA where P = x²y - y³ and Q = xy² + x³. We get ∂Q/∂x = y² + 3x² and ∂P/∂y = x² - 3y², so ∂Q/∂x - ∂P/∂y = y² + 3x² - x² + 3y² = 2x² + 4y². This integrand has no odd symmetry properties that would make it zero. The symmetry about the y-axis doesn't help eliminate terms since 2x² + 4y² is even in x. Direct application of Green's theorem is most efficient.

Question 7

Consider the surface integral $\iint_S \mathbf{F} \cdot d\mathbf{S}$ where $\mathbf{F}(x,y,z) = (yz^2, xz^2, x^2y + y^2z)$ and $S$ is the closed surface consisting of the hemisphere $x^2 + y^2 + z^2 = 4, z \geq 0$ together with the disk $x^2 + y^2 \leq 4, z = 0$ . Which symmetry observation leads to the most efficient evaluation method?

The vector field has rotational symmetry about the z-axis, allowing reduction to a one-dimensional integral
Reflection symmetry about the xz-plane makes certain components of the flux integral vanish on the hemisphere
The divergence theorem applies directly, and $\nabla \cdot \mathbf{F}$ has symmetries that simplify the volume integral (correct answer)
Symmetry about the yz-plane eliminates the contribution from the hemisphere, leaving only the disk contribution

Explanation: Computing ∇·F = ∂(yz²)/∂x + ∂(xz²)/∂y + ∂(x²y + y²z)/∂z = 0 + 0 + y². By the divergence theorem, the surface integral equals ∭ y² dV over the hemisphere. Due to symmetry about the xz-plane, we can compute this as 2∭ y² dV over the part where y ≥ 0, or use spherical coordinates directly. Option A is wrong because F doesn't have full rotational symmetry. Option B incorrectly identifies which symmetries affect which components. Option D is incorrect because symmetry doesn't eliminate the hemisphere's contribution entirely.