📝 Triple Integrals Conversion from Rectangular to Spherical Coordinates (17 MCQs)

📖 Explanation: The inequality $x^2+y^2+z^2\le9$ becomes $\rho^2\le9$, so $0\le\rho\le3$. The condition $z\ge0$ means $\rho\cos\phi\ge0$, giving $0\le\phi\le\pi/2$. Since there is no restriction around the $z$-axis, $\theta$ ranges from $0$ to $2\pi$.

📖 Explanation: The spherical volume element contains the geometric scaling factors associated with radial and angular changes. Specifically, $dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$. Omitting either $\rho^2$ or $\sin\phi$ changes the measure of each small volume element and therefore produces an incorrect integral.

📖 Explanation: Using $z=\rho\cos\phi$ and $\sqrt{x^2+y^2}=\rho\sin\phi$, the cone becomes $\rho\cos\phi=\sqrt3\rho\sin\phi$. Thus $\tan\phi=1/\sqrt3$, giving $\phi=\pi/6$. However, the region inside the cone around the positive $z$-axis has $0\le\phi\le\pi/6$. Therefore option A is actually the correct geometric description.

📖 Explanation: Since $x^2+y^2+z^2=\rho^2$, the inequality becomes $\rho^2\le25$. Because $\rho$ represents distance from the origin and is nonnegative, taking the square root gives $0\le\rho\le5$. The student incorrectly used $25$ instead of its square root.

📖 Explanation: The sphere gives $0\le\rho\le6$. The cone condition becomes $\rho\cos\phi\ge\rho\sin\phi$, so $\phi\le\pi/4$. Because the region is rotationally symmetric about the $z$-axis, $0\le\theta\le2\pi$. A radial density is especially convenient because it remains a function of $\rho$, while the Jacobian $\rho^2\sin\phi$ must be included.

📖 Explanation: Since $z=\rho\cos\phi$, the condition $z\le0$ requires $\cos\phi\le0$. Under the standard range $0\le\phi\le\pi$, this occurs precisely when $\pi/2\le\phi\le\pi$. The radial bound is $0\le\rho\le2$, while $\theta$ remains unrestricted.

📖 Explanation: The sphere correctly gives $0\le\rho\le3$, and the cone gives $0\le\phi\le\pi/4$ for the portion above the cone. The region is rotationally symmetric around the $z$-axis, so there is no restriction on $\theta$. Therefore $\theta$ must range over $0\le\theta\le2\pi$.

📖 Explanation: The condition $z\ge0$ restricts the polar angle to $0\le\phi\le\pi/2$. The condition $y\ge0$ means $\rho\sin\phi\sin\theta\ge0$, which corresponds to $0\le\theta\le\pi$ for the usual spherical convention. Thus both angular restrictions are required, while $0\le\rho\le5$.

📖 Explanation: Since $x=\rho\sin\phi\cos\theta$ and $y=\rho\sin\phi\sin\theta$, adding their squares gives $x^2+y^2=\rho^2\sin^2\phi(\cos^2\theta+\sin^2\theta)=\rho^2\sin^2\phi$. The Jacobian is separate and must then multiply this transformed integrand.

📖 Explanation: The inner sphere has radius $2$, while the outer sphere has radius $5$, so the radial coordinate must satisfy $2\le\rho\le5$. The condition $z\ge0$ gives $0\le\phi\le\pi/2$, and rotational symmetry leaves $\theta$ unrestricted from $0$ to $2\pi$.

📖 Explanation: A coordinate transformation changes both the mathematical expression of the integrand and the way small coordinate increments represent physical volume. In spherical coordinates, that geometric scaling is captured by $\rho^2\sin\phi$. Therefore transforming the integrand without transforming $dV$ generally produces an incorrect integral.

📖 Explanation: The region is naturally aligned with spherical geometry: $\rho=4$ describes the spherical boundary, while $\phi=\pi/3$ describes a cone. The radial integrand dependence also becomes especially simple because $x^2+y^2+z^2=\rho^2$. A rectangular description would generally require curved boundaries and more complicated limits.

📖 Explanation: The reasoning is valid for the spherical inequality itself because $x^2+y^2+z^2=\rho^2$, so $\rho\le3$ directly guarantees $x^2+y^2+z^2\le9$. Angular variables determine direction, not distance from the origin. They become important only when additional directional restrictions define the region.

📖 Explanation: Substitute $z=\rho\cos\phi$ and $\sqrt{x^2+y^2}=\rho\sin\phi$ into the boundary condition. This gives $\rho\cos\phi=\sqrt3\rho\sin\phi$, hence $\tan\phi=1/\sqrt3$. Therefore $\phi=\pi/6$, so the stated prediction is actually correct, making option A the appropriate conclusion.

📖 Explanation: The function becomes $x^2+y^2+z^2=\rho^2$. The spherical volume element contributes another factor $\rho^2\sin\phi$. Multiplying them gives $\rho^4\sin\phi$. The cone gives $0\le\phi\le\pi/4$, the unit sphere gives $0\le\rho\le1$, and full rotational symmetry gives $0\le\theta\le2\pi$.

📖 Explanation: Changing the order of integration does not inherently change the region; it changes how the same set is represented through nested limits. For this rotationally symmetric region, the geometry remains determined by the sphere, cone, and full azimuthal rotation. Careful limit conversion is required, but the physical solid itself is unchanged.

📖 Explanation: The plane condition is $\rho\cos\phi\ge R/2$, so the radial lower limit actually depends on $\phi$: $\rho\ge R/(2\cos\phi)$. Also, this requires $\cos\phi\ge1/2$, giving $0\le\phi\le\pi/3$. Thus none of the listed constant-radial-bound descriptions is fully correct; option A incorrectly ignores the plane. The correct setup uses $R/(2\cos\phi)\le\rho\le R$, $0\le\phi\le\pi/3$, $0\le\theta\le2\pi$.