Using Symmetry - Multivariable Calculus

1. 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?
   Answer: $\iiint_E (x^2 + y^2 + z^2)^2 \, dV$ because the integrand can be factored as the square of the radius 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.
2. 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?
   Answer: The divergence theorem applies directly, and $\nabla \cdot \mathbf{F}$ has symmetries that simplify the volume integral 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.
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?
   Answer: The vector field $\mathbf{F}$ is conservative, making the integral zero, but this isn't due to the symmetry of the region 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.
4. 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?
   Answer: Only the first term $x^3\sin(y^2)$ integrates to zero; the other terms contribute non-zero values 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.
5. 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?
   Answer: Convert to elliptical coordinates and recognize that the odd symmetry in each variable makes the integral zero 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.
6. 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?
   Answer: Symmetry arguments partially apply; some terms can be simplified but the analysis is more complex than standard symmetry cases 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.
7. 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?
   Answer: Apply Green's theorem directly since the partial derivatives create a simple integrand over the semicircular region 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.

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