Change of Variables - Multivariable Calculus

A change of variables $T(u,v) = (x,y)$ maps a region $S$ in the $uv$ -plane to a region $R$ in the $xy$ -plane. If the Jacobian determinant of this transformation is $J(u,v) = u^2+1$ and the area of $R$ is 10, what can be concluded about the integral $\iint_S (u^2+1) \, du \, dv$ ?
Answer: The integral is equal to 10. When you encounter a change of variables problem involving a Jacobian determinant, you're dealing with how areas transform between coordinate systems. The key relationship is that the Jacobian tells you the local scaling factor for area elements. The fundamental theorem for change of variables states that $\iint_R f(x,y) \, dx \, dy = \iint_S f(T(u,v)) |J(u,v)| \, du \, dv$ , where $J(u,v)$ is the Jacobian determinant. More specifically, the area of region $R$ equals $\text{Area}(R) = \iint_S |J(u,v)| \, du \, dv$ . Since the Jacobian is $J(u,v) = u^2 + 1$ , which is always positive (as $u^2 \geq 0$ ), we have $|J(u,v)| = u^2 + 1$ . Therefore: $\text{Area}(R) = \iint_S (u^2 + 1) \, du \, dv = 10$ This means the integral $\iint_S (u^2+1) \, du \, dv = 10$ . Option A is wrong because the relationship between the Jacobian and area transformation gives us exactly what we need to determine the integral's value. Option B incorrectly suggests taking the reciprocal of the area, which has no basis in the change of variables formula. Option C confuses the integral with the area of the original region $S$ , but the integral $\iint_S (u^2+1) \, du \, dv$ actually gives the area of the transformed region $R$ . Remember: when the Jacobian appears as an integrand over the original region, it always equals the area of the transformed region. This is the geometric meaning of the Jacobian determinant.
When using a transformation $x=g(u,v), y=h(u,v)$ to change variables in a double integral, the area element $dA = dx\,dy$ is replaced by $|J(u,v)| \, du \, dv$ . What is the geometric interpretation of the factor $|J(u,v)|$ ?
Answer: The local scaling factor for areas, transforming an infinitesimal area in the $uv$ -plane to the corresponding area in the $xy$ -plane. The Jacobian determinant $J(u,v)$ represents how the transformation $T(u,v)=(x(u,v), y(u,v))$ locally scales areas. An infinitesimal rectangle in the $uv$ -plane with area $\Delta u \Delta v$ is mapped to an infinitesimal parallelogram in the $xy$ -plane with area approximately $|J(u,v)| \Delta u \Delta v$ . Therefore, $|J(u,v)|$ is the local scaling factor for areas from the $uv$ -plane to the $xy$ -plane. Choice A describes $1/|J(u,v)|$ . Choice C confuses the Jacobian with the Hessian matrix. Choice D is not a standard interpretation of the Jacobian.
To compute $\iint_R (x^2+y^2) \, dA$ where $R$ is the square with vertices $(1,0), (0,1), (-1,0), (0,-1)$ , which substitution is most effective?
Answer: $u = x+y$ , $v = x-y$ When evaluating double integrals over regions with unusual boundaries, the key is choosing a substitution that transforms the region into something with simpler limits of integration, like a rectangle or standard geometric shape. The substitution $u = x+y$ , $v = x-y$ is most effective because it transforms the diamond-shaped region into a rectangle. To see why, note that the vertices $(1,0), (0,1), (-1,0), (0,-1)$ become $(1,1), (1,-1), (-1,-1), (-1,1)$ respectively under this transformation. The diamond's boundary lines $x+y = 1$ , $x+y = -1$ , $x-y = 1$ , and $x-y = -1$ become the simple rectangular boundaries $u = 1$ , $u = -1$ , $v = 1$ , and $v = -1$ . With the Jacobian $J = 1/2$ , the integral becomes $\frac{1}{2}\int_{-1}^{1}\int_{-1}^{1}(u^2+v^2) \, du \, dv$ , which is straightforward to evaluate. Choice A ( $u = x^2$ , $v = y^2$ ) doesn't simplify the region's boundaries meaningfully. Choice B (polar coordinates) might seem natural for $x^2+y^2$ , but the diamond region creates complicated limits in polar form since it's not centered at the origin with circular symmetry. Choice C ( $u = x$ , $v = y/\sqrt{1-x^2}$ ) appears to be a trigonometric substitution that's inappropriate for this region and integrand. Strategy tip: When the region has linear boundaries, look for linear substitutions that align with those boundaries. Diagonal regions often benefit from rotational transformations like $u = x+y$ , $v = x-y$ .
To evaluate $\iint_R \sqrt{x^2 + y^2} \, e^{x^2 + y^2} \, dA$ where $R$ is the region $x^2 + y^2 \leq 4$ with $x \geq 0$ and $y \geq x$ , which setup correctly describes the integral in polar coordinates?
Answer: $\int_{\pi/4}^{\pi/2} \int_0^2 r \cdot r e^{r^2} \, dr \, d\theta$ In polar coordinates, $\sqrt{x^2 + y^2} = r$ and $x^2 + y^2 = r^2$ , so the integrand becomes $r e^{r^2}$ . Including the Jacobian $r$ , we get $r \cdot r e^{r^2} = r^2 e^{r^2}$ . The region constraints are: $x^2 + y^2 \leq 4$ gives $r \leq 2$ ; $x \geq 0$ means we're in the right half-plane; $y \geq x$ means $\tan\theta \geq 1$ , so $\theta \geq \frac{\pi}{4}$ . Combined with $x \geq 0$ , we need $\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$ . Choice A uses $0 \leq \theta \leq \frac{\pi}{4}$ , which gives the region where $0 \leq y \leq x$ . Choice C includes the entire right half-plane. Choice D extends into the second quadrant where $x < 0$ .
To evaluate the integral $\iint_R \frac{y}{x} \, dA$ over the region $R$ in the first quadrant bounded by the curves $y=x$ , $y=3x$ , $xy=1$ , and $xy=2$ , the substitution $u = y/x$ and $v = xy$ is used. What is the value of the integral?
Answer: $1$ The transformation is $u=y/x$ and $v=xy$ . The region $R$ is bounded by $y/x=1$ , $y/x=3$ , $xy=1$ , and $xy=2$ . In the $uv$ -plane, this corresponds to the rectangular region $S$ defined by $1 \le u \le 3$ and $1 \le v \le 2$ . The integrand becomes $f(x,y) = y/x = u$ . We must find the Jacobian of the transformation. It is easier to compute the Jacobian of the inverse transformation first: $J^{-1} = \frac{\partial(u,v)}{\partial(x,y)} = \det \begin{pmatrix} -y/x^2 & 1/x \\ y & x \end{pmatrix} = (-y/x) - (y/x) = -2y/x = -2u$ . The Jacobian determinant is $J = \frac{\partial(x,y)}{\partial(u,v)} = \frac{1}{J^{-1}} = -\frac{1}{2u}$ . The area element transforms as $dA = |J| \, du \, dv = |-\frac{1}{2u}| \, du \, dv = \frac{1}{2u} \, du \, dv$ (since $u=y/x > 0$ in the first quadrant). The transformed integral is $\int_1^2 \int_1^3 u \cdot \frac{1}{2u} \, du \, dv = \int_1^2 \int_1^3 \frac{1}{2} \, du \, dv$ . Evaluating this gives $\frac{1}{2} \cdot (3-1) \cdot (2-1) = \frac{1}{2} \cdot 2 \cdot 1 = 1$ .
Consider the elliptic integral $\iint_R \frac{1}{\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}} \, dA$ where $R$ is the region $\frac{x^2}{a^2} + \frac{y^2}{b^2} \leq \frac{1}{4}$ . Which substitution transforms this into the most manageable form?
Answer: $x = \frac{a}{2}r\cos\theta, y = \frac{b}{2}r\sin\theta$ with $0 \leq r \leq 1$ The substitution $x = \frac{a}{2}r\cos\theta, y = \frac{b}{2}r\sin\theta$ transforms the elliptical region $\frac{x^2}{a^2} + \frac{y^2}{b^2} \leq \frac{1}{4}$ into $r^2 \leq 1$ , and the integrand becomes $\frac{1}{\sqrt{1-\frac{r^2}{4}}} = \frac{2}{\sqrt{4-r^2}}$ . The Jacobian is $\frac{ab}{4}r$ . This gives a standard form integral in polar coordinates. Choice A uses Cartesian substitution but doesn't utilize the circular symmetry. Choice B has incorrect variable pairing. Choice D changes to a triangular-like region but makes the integrand more complex rather than simpler.
The integral $\iint_R x \, dA$ is transformed using the substitution $x=u+v, y=u-v$ into the integral $\int_0^2 \int_0^1 (u+v) \cdot 2 \, du \, dv$ . What is the region of integration $R$ in the $xy$ -plane?
Answer: The parallelogram with vertices $(0,0), (1,1), (2,-2), (3,-1)$ . The transformed integral has limits $0 \le u \le 1$ and $0 \le v \le 2$ . This defines a rectangular region $S$ in the $uv$ -plane with vertices $(0,0), (1,0), (0,2),$ and $(1,2)$ . The original region $R$ is the image of $S$ under the transformation $T(u,v)=(u+v, u-v)$ . We find the vertices of $R$ by transforming the vertices of $S$ : $T(0,0)=(0,0)$ ; $T(1,0)=(1,1)$ ; $T(0,2)=(2,-2)$ ; $T(1,2)=(1+2, 1-2)=(3,-1)$ . Thus, $R$ is the parallelogram with these four vertices.
What is the area of the region $R$ in the $xy$ -plane defined by the inequality $x^2 - 2xy + 5y^2 \le 1$ ?
Answer: $\pi/2$ The expression $x^2 - 2xy + 5y^2$ can be rewritten by completing the square: $x^2 - 2xy + y^2 + 4y^2 = (x-y)^2 + (2y)^2$ . Let's use the substitution $u = x-y$ and $v = 2y$ . The region $R$ in the $xy$ -plane is transformed into the region $S$ in the $uv$ -plane defined by $u^2 + v^2 \le 1$ , which is a unit disk. The area of $R$ is $\iint_R 1 \, dA$ . To use the change of variables, we need the Jacobian determinant $|J| = |\frac{\partial(x,y)}{\partial(u,v)}|$ . First, we solve for $x$ and $y$ : $y = v/2$ and $x = u+y = u+v/2$ . The Jacobian is $J = \det \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} = \det \begin{pmatrix} 1 & 1/2 \\ 0 & 1/2 \end{pmatrix} = 1/2$ . The area of $R$ is $\iint_S |1/2| \, du \, dv = \frac{1}{2} \iint_S 1 \, du \, dv = \frac{1}{2} \cdot \text{Area}(S)$ . Since $S$ is a unit disk, its area is $\pi(1)^2 = \pi$ . Therefore, the area of $R$ is $\pi/2$ .
What is the area of the region bounded by the ellipses $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and $\frac{x^2}{36} + \frac{y^2}{16} = 1$ ?
Answer: $18\pi$ When you encounter a problem asking for the area between two concentric ellipses, you're finding the difference between their individual areas. Both ellipses here are centered at the origin, with the second ellipse being larger than the first. For any ellipse in the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ , the area is $\pi ab$ . Let's calculate each area: The inner ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ has $a^2 = 9$ and $b^2 = 4$ , so $a = 3$ and $b = 2$ . Its area is $\pi(3)(2) = 6\pi$ . The outer ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$ has $a^2 = 36$ and $b^2 = 16$ , so $a = 6$ and $b = 4$ . Its area is $\pi(6)(4) = 24\pi$ . The area between the ellipses is $24\pi - 6\pi = 18\pi$ , which is answer B. Let's examine the wrong answers: A ( $6\pi$ ) is just the area of the inner ellipse alone—this misses that we need the region between the curves. C ( $15\pi$ ) might result from calculation errors when finding the semi-axes. D ( $24\pi$ ) is the area of the outer ellipse alone, ignoring that we must subtract the inner area. Study tip: For "area between curves" problems, always subtract the inner area from the outer area. Double-check that you're using the ellipse area formula $\pi ab$ correctly by taking square roots of the denominators to find the semi-axes.
The region $R$ is defined by $1 \leq x^2 + y^2 \leq 4$ and $\frac{\sqrt{3}}{3} \leq \frac{y}{x} \leq \sqrt{3}$ with $x > 0, y > 0$ . To evaluate $\iint_R \frac{x^2 - y^2}{x^2 + y^2} \, dA$ , which substitution makes both the integrand and region bounds simplest?
Answer: $x = r\cos\theta, y = r\sin\theta$ with $1 \leq r \leq 2$ and $\frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}$ The region is an annular sector, ideal for polar coordinates. We have $1 \leq x^2 + y^2 \leq 4$ becomes $1 \leq r^2 \leq 4$ , so $1 \leq r \leq 2$ . For the angular bounds, $\frac{y}{x} = \tan\theta$ . Since $\frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} = \tan(\frac{\pi}{6})$ and $\sqrt{3} = \tan(\frac{\pi}{3})$ , we get $\frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}$ . The integrand becomes $\frac{r^2\cos^2\theta - r^2\sin^2\theta}{r^2} = \cos^2\theta - \sin^2\theta = \cos(2\theta)$ . Choice B has the angle bounds reversed. Choice C works but is more complex than needed. Choice D incorrectly uses $r^2$ bounds instead of $r$ bounds.
To evaluate $\iint_R (x-2y) \, dA$ where $R$ is the triangle with vertices $(0,0), (2,1), (1,3)$ , a linear transformation $T(u,v) = (x(u,v), y(u,v))$ is used to map the standard triangle $S$ in the $uv$ -plane with vertices $(0,0), (1,0), (0,1)$ to $R$ . Which of the following is the resulting integral over $S$ ?
Answer: $\int_0^1 \int_0^{1-u} (-25v) \, dv \, du$ First, find the linear transformation $x = au+bv$ , $y=cu+dv$ . It must map the vertices of $S$ to the vertices of $R$ . $T(0,0)=(0,0)$ is satisfied. $T(1,0)=(a,c)=(2,1)$ , so $a=2, c=1$ . $T(0,1)=(b,d)=(1,3)$ , so $b=1, d=3$ . The transformation is $x=2u+v, y=u+3v$ . The Jacobian is $J = \det \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} = 6-1=5$ . The integrand $x-2y$ becomes $(2u+v) - 2(u+3v) = 2u+v-2u-6v = -5v$ . The area element is $dA = |J| \, du \, dv = 5 \, du \, dv$ . The region $S$ is described by $u \ge 0, v \ge 0, u+v \le 1$ . This gives the integral limits $0 \le u \le 1$ and $0 \le v \le 1-u$ . The transformed integral is $\iint_S (-5v) \cdot 5 \, du \, dv = \int_0^1 \int_0^{1-u} (-25v) \, dv \, du$ .
Consider the change of variables $u = 2y - 3x$ and $v = x + y$ . Which expression correctly represents the area element $dA = dx \, dy$ in terms of $du \, dv$ ?
Answer: $\frac{1}{5} \, du \, dv$ The area element transforms according to $dx \, dy = |J| \, du \, dv$ , where $J = \frac{\partial(x,y)}{\partial(u,v)}$ . It is often easier to first compute $J^{-1} = \frac{\partial(u,v)}{\partial(x,y)}$ and then take the reciprocal. Here, $J^{-1} = \det \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix} = \det \begin{pmatrix} -3 & 2 \\ 1 & 1 \end{pmatrix} = (-3)(1) - (2)(1) = -5$ . Then $J = \frac{1}{J^{-1}} = -\frac{1}{5}$ . The area element is $dA = |J| \, du \, dv = |-\frac{1}{5}| \, du \, dv = \frac{1}{5} \, du \, dv$ .
To evaluate $\iint_R (x-1) \, dA$ over the disk $R$ defined by $(x-1)^2 + y^2 \le 4$ , a student uses the substitution $x=1+r\cos\theta$ and $y=r\sin\theta$ . Which of the following is the correct setup for the transformed integral?
Answer: $\int_0^{2\pi} \int_0^2 r^2 \cos\theta \, dr \, d\theta$ The substitution is a shifted version of polar coordinates. The region $R$ is a disk of radius 2 centered at $(1,0)$ . In the new coordinates, the boundary $(x-1)^2 + y^2 = 4$ becomes $(r\cos\theta)^2 + (r\sin\theta)^2 = 4$ , which simplifies to $r^2=4$ , or $r=2$ . So, the region of integration is $0 \le r \le 2$ and $0 \le \theta \le 2\pi$ . The integrand is $x-1 = (1+r\cos\theta) - 1 = r\cos\theta$ . The Jacobian for this transformation is $J = \det \begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix} = r$ . Thus, $dA = r \, dr \, d\theta$ . The integral becomes $\int_0^{2\pi} \int_0^2 (r\cos\theta) \cdot r \, dr \, d\theta = \int_0^{2\pi} \int_0^2 r^2 \cos\theta \, dr \, d\theta$ .
For the integral $\iiint_E (x^2 + y^2) \, dV$ where $E$ is the solid bounded by $z = x^2 + y^2$ and $z = 8 - x^2 - y^2$ , which coordinate system and bounds correctly describe the region?
Answer: Cylindrical: $0 \leq r \leq 2$ , $r^2 \leq z \leq 8 - r^2$ , $0 \leq \theta \leq 2\pi$ When evaluating triple integrals over regions bounded by surfaces, your first step is identifying the most natural coordinate system. Here, both bounding surfaces $z = x^2 + y^2$ and $z = 8 - x^2 - y^2$ involve $x^2 + y^2$ , which immediately suggests cylindrical coordinates where $r^2 = x^2 + y^2$ . To find the correct bounds, first determine where these surfaces intersect. Setting $x^2 + y^2 = 8 - x^2 - y^2$ gives $2(x^2 + y^2) = 8$ , so $x^2 + y^2 = 4$ . In cylindrical coordinates, this means $r^2 = 4$ , so $r = 2$ . For any point $(r,\theta)$ with $0 \leq r \leq 2$ , the region extends vertically from the lower surface $z = r^2$ to the upper surface $z = 8 - r^2$ . The angle $\theta$ covers the full circle: $0 \leq \theta \leq 2\pi$ . Answer choice A incorrectly sets the upper $z$ -bound as $8 - 2r^2$ instead of $8 - r^2$ . This stems from confusing the cylindrical relationship $r^2 = x^2 + y^2$ . Answer choice B uses the wrong radial bound $r \leq \sqrt{8}$ . While the surfaces intersect when $x^2 + y^2 = 4$ , the maximum radius is $r = 2$ , not $\sqrt{8}\$. Answer choice C attempts spherical coordinates, which is unnecessarily complicated for this geometry since the surfaces are naturally expressed in terms of$ $x^2$ + $y^2$ $. **Study tip**: When you see$ $x^2$ + $y^2$ $$ in surface equations, immediately consider cylindrical coordinates. Always find intersection curves first to determine your bounds.
Consider the transformation $T: u = 2x + y, v = x - y$ applied to the triangular region with vertices at $(0,0)$ , $(2,0)$ , and $(1,2)$ . What is the area of the transformed region in the $uv$ -plane?
Answer: $6$ First, find the Jacobian of the transformation. We have $\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} = -2 - 1 = -3$ , so $|J| = 3$ . The original triangle has vertices at $(0,0)$ , $(2,0)$ , and $(1,2)$ . Using the shoelace formula, the area is $\frac{1}{2}|0(0-2) + 2(2-0) + 1(0-0)| = \frac{1}{2} \cdot 4 = 2$ . Under the transformation, areas are multiplied by $|J| = 3$ , so the transformed area is $2 \times 3 = 6$ . Choice A gives the original area. Choice B gives just the Jacobian magnitude. Choice D squares the Jacobian incorrectly.
To evaluate $\iint_R e^{\frac{x-y}{x+y}} \, dA$ over the region $R$ bounded by $x + y = 1$ , $x + y = 4$ , $x - y = -1$ , and $x - y = 2$ , which change of variables simplifies the integrand most effectively?
Answer: $u = x + y, v = x - y$ with bounds $1 \leq u \leq 4$ and $-1 \leq v \leq 2$ The integrand $e^{\frac{x-y}{x+y}}$ suggests using $u = x + y$ and $v = x - y$ so that the exponent becomes $\frac{v}{u}$ . The region boundaries become the lines $u = 1, u = 4, v = -1, v = 2$ , forming a rectangle in the $uv$ -plane. Choice B reverses the variables but gives the same transformation (just with $u$ and $v$ swapped). Choice C makes the integrand simpler but complicates the region. Choice D (polar coordinates) doesn't help with this particular integrand structure and makes the rectangular region unnecessarily complex.
To evaluate $\iint_R (x^2 + y^2) \, dA$ where $R$ is the region bounded by $x^2 + y^2 = 4$ , $x^2 + y^2 = 9$ , $y = x$ , and $y = \sqrt{3}x$ in the first quadrant, which substitution and integration bounds are most appropriate?
Answer: $x = r\cos\theta, y = r\sin\theta$ with $2 \leq r \leq 3$ and $\frac{\pi}{4} \leq \theta \leq \frac{\pi}{3}$ The region is bounded by two circles and two lines through the origin. Polar coordinates are ideal here. The circles $x^2 + y^2 = 4$ and $x^2 + y^2 = 9$ become $r = 2$ and $r = 3$ . The line $y = x$ has slope 1, so $\tan\theta = 1$ giving $\theta = \frac{\pi}{4}$ . The line $y = \sqrt{3}x$ has slope $\sqrt{3}$ , so $\tan\theta = \sqrt{3}$ giving $\theta = \frac{\pi}{3}$ . Choice B uses $r^2$ values instead of $r$ values. Choice C reverses the angle bounds. Choice D uses a valid substitution but is unnecessarily complex for this geometry.
Consider the transformation $T$ given by $u = x + 2y$ , $v = 3x - y$ . If this transformation maps the unit square $[0,1] \times [0,1]$ to a region $R$ in the $uv$ -plane, what is the area of $R$ ?
Answer: $7$ The Jacobian of the transformation is $\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} = (1)(-1) - (2)(3) = -1 - 6 = -7$ . The area scaling factor is $|J| = 7$ . Since the original unit square has area 1, the transformed region has area $1 \times 7 = 7$ . Choice A gives $\frac{1}{|J|}$ instead of $|J|$ . Choice B gives the original area without transformation. Choice C incorrectly computes the Jacobian as $1 + 2 + 3 + 1 = 7$ instead of the determinant.
For the integral $\iiint_E z \, dV$ where $E$ is the region inside both $x^2 + y^2 + z^2 = 9$ and $x^2 + y^2 \leq z^2$ , which coordinate system and bounds are most appropriate?
Answer: Spherical: $0 \leq \rho \leq 3$ , $0 \leq \phi \leq \frac{\pi}{4}$ and $\frac{3\pi}{4} \leq \phi \leq \pi$ , $0 \leq \theta \leq 2\pi$ The region is inside a sphere of radius 3 and inside the double cone $x^2 + y^2 = z^2$ . In spherical coordinates, the sphere is $\rho = 3$ and the cone becomes $\rho^2\sin^2\phi = \rho^2\cos^2\phi$ , which simplifies to $\tan\phi = \pm 1$ , giving $\phi = \frac{\pi}{4}$ or $\phi = \frac{3\pi}{4}$ . The region consists of two parts: the upper cone ( $0 \leq \phi \leq \frac{\pi}{4}$ ) and the lower cone ( $\frac{3\pi}{4} \leq \phi \leq \pi$ ). Choice A misses the lower cone. Choice C uses cylindrical coordinates but doesn't properly handle the spherical boundary. Choice D describes the region outside the cones, not inside.

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