Multivariable Calculus Quiz: Sketching Regions And Bounds

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Sketching Regions And BoundsQuestion 1 of 15

The region $R$ is bounded by the curves $xy = 1$ , $xy = 4$ , $y = x$ , and $y = 4x$ . Using the transformation $u = xy$ and $v = y/x$ , what are the bounds for the transformed region in the $uv$ -plane?

1 \leq u \leq 4, \quad 1 \leq v \leq 4

1 \leq u \leq 4, \quad 1/4 \leq v \leq 1

1 \leq u \leq 4, \quad 1 \leq v \leq 2

1/4 \leq u \leq 1, \quad 1 \leq v \leq 4

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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Sketching Regions And Bounds

Practice Sketching Regions And Bounds 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 Sketching Regions And Bounds, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

How to use this quiz

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

$1 \leq u \leq 4, \quad 1 \leq v \leq 4$ (correct answer)
$1 \leq u \leq 4, \quad 1/4 \leq v \leq 1$
$1 \leq u \leq 4, \quad 1 \leq v \leq 2$
$1/4 \leq u \leq 1, \quad 1 \leq v \leq 4$

Explanation: Under the transformation

u = xy

and

v = y/x

, let's see how each boundary curve transforms: The curves

xy = 1

and

xy = 4

become

u = 1

and

u = 4

respectively. The curve

y = x

gives

v = y/x = x/x = 1

. The curve

y = 4x

gives

v = y/x = 4x/x = 4

. Therefore, the transformed region is

1 \leq u \leq 4

and

1 \leq v \leq 4

. Choice B incorrectly computes the

v

bounds by taking reciprocals. Choice C incorrectly limits the upper

v

bound. Choice D swaps the

u

and

v

ranges incorrectly.

Question 2

Let $R$ be the region in the $xy$ -plane that lies inside the cardioid $r = 1 + \cos\theta$ and outside the circle $r=1$ . Which iterated integral in polar coordinates gives the volume of the solid under the surface $z=y$ and above the region $R$ ?

$\int_{-\pi/2}^{\pi/2} \int_1^{1+\cos\theta} r^2 \sin\theta \, dr \, d\theta$ (correct answer)
$\int_{-\pi/2}^{\pi/2} \int_1^{1+\cos\theta} r \sin\theta \, dr \, d\theta$
$\int_{0}^{2\pi} \int_1^{1+\cos\theta} r^2 \sin\theta \, dr \, d\theta$
$\int_{-\pi/2}^{\pi/2} \int_1^{1+\cos\theta} r^2 \cos\theta \, dr \, d\theta$

Explanation: The region of integration

R

is defined by the bounds of

r

and

\theta

. The curves intersect when

1 = 1 + \cos\theta

, which means

\cos\theta = 0

. This occurs at

\theta = -\pi/2

and

\theta = \pi/2

. So, the bounds for

\theta

are from

-\pi/2

\pi/2

. For any such

\theta

r

ranges from the inner circle

r=1

to the outer cardioid

r=1+\cos\theta

. The volume is given by the double integral of the height function

z=y

over

R

. In polar coordinates,

y=r\sin\theta

, and the area element is

dA = r \, dr \, d\theta

. The integrand is thus

(r\sin\theta) \cdot r = r^2\sin\theta

. Combining these gives the integral

\int_{-\pi/2}^{\pi/2} \int_1^{1+\cos\theta} r^2 \sin\theta \, dr \, d\theta

Question 3

Let $R$ be the trapezoidal region in the $xy$ -plane with vertices $(1,0), (2,0), (0,2), (0,1)$ . Using the transformation $u=x+y$ and $v=y-x$ , the integral $\iint_R (x+y) \, dA$ is transformed into which of the following integrals over a region $S$ in the $uv$ -plane?

$\int_1^2 \int_{-u}^{u} \frac{1}{2} u \, dv \, du$ (correct answer)
$\int_1^2 \int_{u-2}^{2-u} 2 u \, dv \, du$
$\int_1^2 \int_{-u}^{u} u \, dv \, du$
$\int_0^2 \int_{1-v}^{2-v} \frac{1}{2} u \, du \, dv$

Explanation: First, we transform the boundaries of

R

. The vertices define the lines

y=0

x=0

x+y=1

, and

x+y=2

. In terms of

u

and

v

x+y=u

means two boundaries are

u=1

and

u=2

. The line

y=0

means

u=x, v=-x

, so

v=-u

. The line

x=0

means

u=y, v=y

, so

v=u

. The region

S

in the

uv

-plane is bounded by

u=1, u=2, v=-u, v=u

. Next, we find the Jacobian of the transformation from

(u,v)

(x,y)

. We need

x

and

y

in terms of

u

and

v

. Adding the transformations gives

u+v=2y \implies y=(u+v)/2

. Subtracting gives

u-v=2x \implies x=(u-v)/2

. The Jacobian determinant is

J = \det \begin{pmatrix} \partial x/\partial u & \partial x/\partial v \\ \partial y/\partial u & \partial y/\partial v \end{pmatrix} = \det \begin{pmatrix} 1/2 & -1/2 \\ 1/2 & 1/2 \end{pmatrix} = 1/4 - (-1/4) = 1/2

. The integrand

(x+y)

becomes

u

. So the integral is

\iint_S u |J| \, dA_{uv} = \int_1^2 \int_{-u}^{u} u \cdot \frac{1}{2} \, dv \, du

Question 4

Let $R$ be the region in the first quadrant defined by the inequalities $1 \le x^2+y^2 \le 4$ and $y \le x \le \sqrt{3}y$ . Which of the following integrals in polar coordinates represents the area of $R$ ?

$\int_{\pi/6}^{\pi/4} \int_1^4 r \, dr \, d\theta$
$\int_{\pi/4}^{\pi/3} \int_1^2 r \, dr \, d\theta$
$\int_{\pi/6}^{\pi/4} \int_1^2 r \, dr \, d\theta$ (correct answer)
$\int_{\pi/4}^{\pi/3} \int_1^2 1 \, dr \, d\theta$

Explanation: When converting a region from Cartesian to polar coordinates, you need to carefully translate each boundary condition and remember that area integrals in polar coordinates always include the Jacobian factor

r

. Let's analyze each boundary of region

R

. The condition

1 \le x^2+y^2 \le 4

becomes

1 \le r^2 \le 4

, so

1 \le r \le 2

. The linear boundaries

y \le x

and

x \le \sqrt{3}y

require converting to polar form using

x = r\cos\theta

and

y = r\sin\theta

. For

y \le x

r\sin\theta \le r\cos\theta

, which simplifies to

\tan\theta \le 1

, giving us

\theta \le \pi/4

. For

x \le \sqrt{3}y

r\cos\theta \le \sqrt{3}r\sin\theta

, which simplifies to

\cot\theta \le \sqrt{3}

, so

\tan\theta \ge 1/\sqrt{3}

, giving us

\theta \ge \pi/6

. Therefore,

\pi/6 \le \theta \le \pi/4

and

1 \le r \le 2

. The area integral is

\int_{\pi/6}^{\pi/4} \int_1^2 r \, dr \, d\theta

, which is choice C. Choice A has the wrong

r

limits (4 instead of 2). Choice B uses incorrect

\theta

limits (

\pi/4

\pi/3

instead of

\pi/6

\pi/4

). Choice D is missing the crucial Jacobian factor

r

in the integrand. Key strategy: Always sketch the region first, then systematically convert each boundary. Remember that area in polar coordinates requires the factor

r

, and double-check your angle conversions using basic trigonometric values.

Question 5

The region $R$ in the $xy$ -plane is defined by $x^2 + y^2 \leq 9$ , $x^2 + y^2 \geq 1$ , and $x + y \geq 0$ . Which of the following polar coordinate setups correctly represents $\iint_R f(x,y) \, dA$ ?

$\int_{-\pi/4}^{3\pi/4} \int_1^3 f(r\cos\theta, r\sin\theta) \, r \, dr \, d\theta$ (correct answer)
$\int_{-3\pi/4}^{\pi/4} \int_1^3 f(r\cos\theta, r\sin\theta) \, r \, dr \, d\theta$
$\int_0^{\pi} \int_1^3 f(r\cos\theta, r\sin\theta) \, r \, dr \, d\theta$
$\int_{\pi/4}^{5\pi/4} \int_1^3 f(r\cos\theta, r\sin\theta) \, r \, dr \, d\theta$

Explanation: The region is an annulus (

1 \leq r \leq 3

) intersected with the half-plane

x + y \geq 0

. In polar coordinates,

x + y \geq 0

becomes

r\cos\theta + r\sin\theta \geq 0

, which simplifies to

r(\cos\theta + \sin\theta) \geq 0

. Since

r > 0

, we need

\cos\theta + \sin\theta \geq 0

. This occurs when

\sin\theta \geq -\cos\theta

, or

\tan\theta \geq -1

. This inequality is satisfied when

\theta \in [-\pi/4 + 2\pi k, 3\pi/4 + 2\pi k]

for integer

k

. Taking the principal range, we get

\theta \in [-\pi/4, 3\pi/4]

. The radial bounds are simply

1 \leq r \leq 3

. Choice B has the wrong angular range. Choice C includes regions where

x + y < 0

. Choice D has an incorrect angular range that misses part of the region.

Question 6

Consider the region $R$ bounded by the cylinder $x^2 + y^2 = 4$ and the planes $z = 0$ , $z = 2$ , and $x + y + z = 3$ . Using cylindrical coordinates, which integral setup correctly computes the volume of $R$ ?

$\int_0^{2\pi} \int_0^2 \int_0^{\min(2, 3-r\cos\theta-r\sin\theta)} r \, dz \, dr \, d\theta$
$\int_0^{2\pi} \int_0^2 \int_{\max(0, 3-r\cos\theta-r\sin\theta)}^2 r \, dz \, dr \, d\theta$ (correct answer)
$\int_0^{2\pi} \int_0^2 \int_0^2 r \, dz \, dr \, d\theta$
$\int_0^{2\pi} \int_0^2 \int_{3-r\cos\theta-r\sin\theta}^2 r \, dz \, dr \, d\theta$

Explanation: The region is inside the cylinder

x^2 + y^2 = 4

(so

0 \leq r \leq 2

) and bounded by three planes. In cylindrical coordinates:

z = 0

z = 2

, and

x + y + z = 3

becomes

r\cos\theta + r\sin\theta + z = 3

, or

z = 3 - r\cos\theta - r\sin\theta

. The region exists where all constraints are satisfied simultaneously. For the

z

bounds, we need

z \geq 0

z \leq 2

, and the plane constraint. The plane

x + y + z = 3

can either provide an upper bound (when

3 - r\cos\theta - r\sin\theta < 2

) or a lower bound (when

3 - r\cos\theta - r\sin\theta > 0

). The lower bound for

z

\max(0, 3 - r\cos\theta - r\sin\theta)

, and the upper bound is

2

. This is because when

3 - r\cos\theta - r\sin\theta > 0

, the plane is above

z = 0

, so we start from the plane. When

3 - r\cos\theta - r\sin\theta \leq 0

, we start from

z = 0

. Choice A uses min instead of max and has the wrong upper bound. Choice C ignores the plane constraint entirely. Choice D doesn't use the max function and could give negative lower bounds.

Question 7

Let $E$ be the solid region bounded by the paraboloid $z=x^2+y^2$ and the plane $z=2x+3$ . Which of the following describes the projection of $E$ onto the $xy$ -plane?

The disk defined by $x^2+y^2 \le 3$ .
The disk defined by $(x-1)^2+y^2 \le 4$ . (correct answer)
The region bounded by the parabola $y^2=2x+3$ .
The region under the line $y=2x+3$ in the first quadrant.

Explanation: The projection of the solid region

E

onto the

xy

-plane is determined by the intersection of its bounding surfaces. We set the

z

-values equal to find the curve of intersection:

x^2+y^2 = 2x+3

. To identify this curve, we can complete the square for the

x

terms:

x^2 - 2x + y^2 = 3 \implies (x^2 - 2x + 1) + y^2 = 3 + 1 \implies (x-1)^2 + y^2 = 4

. This is the equation of a circle in the

xy

-plane centered at

(1,0)

with a radius of

2

. The projection is the region enclosed by this circle, which is the disk defined by the inequality

(x-1)^2+y^2 \le 4

Question 8

Consider the iterated integral $I = \int_0^1 \int_{\sqrt{y}}^{2-y} f(x,y) \, dx \, dy$ . Which of the following expressions is equivalent to $I$ when the order of integration is reversed?

$\int_0^1 \int_{x^2}^{2-x} f(x,y) \, dy \, dx$
$\int_0^2 \int_0^{x^2} f(x,y) \, dy \, dx$
$\int_0^1 \int_0^{x^2} f(x,y) \, dy \, dx + \int_1^2 \int_0^{2-x} f(x,y) \, dy \, dx$ (correct answer)
$\int_0^1 \int_0^{y^2} f(x,y) \, dy \, dx + \int_1^2 \int_0^{2-y} f(x,y) \, dy \, dx$

Explanation: The region of integration is defined by

0 \le y \le 1

and

\sqrt{y} \le x \le 2-y

. The left boundary is

x=\sqrt{y}

(or

y=x^2

for

x \ge 0

) and the right boundary is

x=2-y

(or

y=2-x

). The intersection of these curves occurs when

x^2 = 2-x

, which gives

x^2+x-2=0

, or

(x+2)(x-1)=0

. Since

x=\sqrt{y} \ge 0

, the intersection is at

x=1

, which corresponds to

y=1

. To reverse the order of integration, we must express the bounds of

y

as functions of

x

. The region must be split at

x=1

. For

0 \le x \le 1

, the region is bounded below by

y=0

and above by the parabola

y=x^2

. For

1 \le x \le 2

, the region is bounded below by

y=0

and above by the line

y=2-x

. This leads to the sum of two integrals:

\int_0^1 \int_0^{x^2} f(x,y) \, dy \, dx + \int_1^2 \int_0^{2-x} f(x,y) \, dy \, dx

Question 9

A solid region $E$ lies above the cone $z = \sqrt{x^2+y^2}$ and below the sphere $x^2+y^2+z^2=z$ . Which of the following integrals in spherical coordinates represents the volume of $E$ ?

$\int_{0}^{2\pi} \int_{0}^{\pi/4} \int_{0}^{1} \rho^2\sin\phi \, d\rho \, d\phi \, d\theta$
$\int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{\cos\phi} \rho^2\sin\phi \, d\rho \, d\phi \, d\theta$
$\int_{0}^{2\pi} \int_{\pi/4}^{\pi/2} \int_{0}^{\cos\phi} \rho^2\sin\phi \, d\rho \, d\phi \, d\theta$
$\int_{0}^{2\pi} \int_{0}^{\pi/4} \int_{0}^{\cos\phi} \rho^2\sin\phi \, d\rho \, d\phi \, d\theta$ (correct answer)

Explanation: First, convert the boundary equations to spherical coordinates. The cone

z = \sqrt{x^2+y^2}

becomes

\rho\cos\phi = \sqrt{(\rho\sin\phi)^2} = \rho\sin\phi

, which simplifies to

\tan\phi=1

, so

\phi=\pi/4

. The region is 'above' the cone, meaning angles are between the

z

-axis and the cone, so

0 \le \phi \le \pi/4

. The sphere

x^2+y^2+z^2=z

becomes

\rho^2 = \rho\cos\phi

, which simplifies to

\rho=\cos\phi

. This defines the outer boundary for

\rho

. The region is symmetric about the

z

-axis, so

0 \le \theta \le 2\pi

. The volume element in spherical coordinates is

dV = \rho^2\sin\phi \, d\rho \, d\phi \, d\theta

. Therefore, the volume is given by the integral

\int_{0}^{2\pi} \int_{0}^{\pi/4} \int_{0}^{\cos\phi} \rho^2\sin\phi \, d\rho \, d\phi \, d\theta

Question 10

The iterated integral $I = \int_0^1 \int_{\arcsin y}^{\pi - \arcsin y} f(x,y) \, dx \, dy$ is taken over a region $R$ . Which of the following accurately describes the region $R$ ?

The region bounded by $y=0$ , $y=1$ , $x=0$ , and $x=\pi$ .
The region bounded by the $x$ -axis and the curve $y=\sin(x)$ for $0 \le x \le \pi$ . (correct answer)
The region bounded by the $y$ -axis and the curve $x=\sin(y)$ for $0 \le y \le \pi$ .
The region in the first quadrant bounded by the circle $x^2+y^2=1$ .

Explanation: The bounds of integration are

0 \le y \le 1

and

\arcsin y \le x \le \pi - \arcsin y

. Let's analyze the bounds for

x

. The left boundary is

x = \arcsin y

, which is equivalent to

y = \sin x

for

x \in [0, \pi/2]

. The right boundary is

x = \pi - \arcsin y

. If we take the sine of both sides,

\sin(x) = \sin(\pi - \arcsin y) = \sin(\arcsin y) = y

. This corresponds to the part of the sine curve where

x \in [\pi/2, \pi]

. Thus, for a fixed

y

between 0 and 1,

x

ranges from the ascending part of the sine curve to the descending part. The overall region

R

is bounded above by

y=\sin x

and below by the

x

-axis (

y=0

) over the interval

x \in [0, \pi]

Question 11

Let $E$ be the solid tetrahedron with vertices at $(0,0,0), (2,0,0), (0,4,0),$ and $(0,0,6)$ . Which of the following iterated integrals gives the volume of $E$ ?

$\int_0^2 \int_0^4 \int_0^6 dz \, dy \, dx$
$\int_0^2 \int_0^{4-2x} \int_0^{6-3x-3y/2} dz \, dy \, dx$ (correct answer)
$\int_0^2 \int_0^{4-2x} \int_0^{12-6x-3y} dz \, dy \, dx$
$\int_0^4 \int_0^{2-y} \int_0^{6-3x-3y/2} dz \, dx \, dy$

Explanation: The four vertices lie on the coordinate planes except for the three axial intercepts

(2,0,0), (0,4,0), (0,0,6)

. The plane passing through these three points has the intercept form equation

\frac{x}{2} + \frac{y}{4} + \frac{z}{6} = 1

. Solving for

z

gives the upper bound for the solid:

z = 6 - 3x - \frac{3}{2}y

. The lower bound is the

xy

-plane,

z=0

. The projection of the tetrahedron onto the

xy

-plane is a triangle with vertices

(0,0), (2,0), (0,4)

. The line connecting

(2,0)

and

(0,4)

is given by

\frac{x}{2} + \frac{y}{4} = 1

, or

y=4-2x

. To integrate with respect to

y

then

x

, the bounds are

0 \le y \le 4-2x

for

0 \le x \le 2

. Combining these gives the integral

\int_0^2 \int_0^{4-2x} \int_0^{6-3x-3y/2} dz \, dy \, dx

Question 12

Let $E$ be the solid region bounded by the parabolic cylinder $y=x^2$ , the plane $y+z=4$ , and the $xy$ -plane ( $z=0$ ). Which of the following iterated integrals represents the volume of $E$ ?

$\int_{-2}^{2} \int_{x^2}^{4} \int_{0}^{4-y} dz \, dy \, dx$ (correct answer)
$\int_{-2}^{2} \int_{0}^{x^2} \int_{0}^{4} dz \, dy \, dx$
$\int_{0}^{4} \int_{-\sqrt{y}}^{\sqrt{y}} \int_{0}^{4} dz \, dx \, dy$
$\int_{-2}^{2} \int_{x^2}^{4} \int_{0}^{4-x^2} dz \, dy \, dx$

Explanation: The solid

E

is bounded below by the

xy

-plane, so

z=0

. It is bounded above by the plane

y+z=4

, so

z=4-y

. Thus, the bounds for

z

are

0 \le z \le 4-y

. To determine the bounds for

x

and

y

, we project the solid onto the

xy

-plane. The projection is the region bounded by

y=x^2

and the intersection of

z=4-y

with

z=0

, which is the line

y=4

. The region is described by

x^2 \le y \le 4

. The intersection of

y=x^2

and

y=4

occurs at

x=\pm 2

. Therefore, the bounds for the projection are

-2 \le x \le 2

and

x^2 \le y \le 4

. Combining these in the order

dz \, dy \, dx

gives the integral

\int_{-2}^{2} \int_{x^2}^{4} \int_{0}^{4-y} dz \, dy \, dx

Question 13

A solid region $E$ is bounded below by the cone $z=\sqrt{x^2+y^2}$ and above by the paraboloid $z=2-x^2-y^2$ . Which of the following integrals in cylindrical coordinates represents the volume of $E$ ?

$\int_{0}^{2\pi} \int_{0}^{1} \int_{r}^{2-r^2} dz \, dr \, d\theta$
$\int_{0}^{2\pi} \int_{0}^{2} \int_{r}^{2-r^2} r \, dz \, dr \, d\theta$
$\int_{0}^{2\pi} \int_{0}^{1} \int_{2-r^2}^{r} r \, dz \, dr \, d\theta$
$\int_{0}^{2\pi} \int_{0}^{1} \int_{r}^{2-r^2} r \, dz \, dr \, d\theta$ (correct answer)

Explanation: In cylindrical coordinates, the cone

z=\sqrt{x^2+y^2}

becomes

z=r

. The paraboloid

z=2-x^2-y^2

becomes

z=2-r^2

. The solid is bounded below by

z=r

and above by

z=2-r^2

. To find the bounds for

r

, we find the intersection of these surfaces:

r = 2-r^2 \implies r^2+r-2=0 \implies (r+2)(r-1)=0

. Since radius

r

must be non-negative,

r=1

. This means the projection of the solid onto the

xy

-plane is a disk of radius 1. So,

0 \le r \le 1

. The solid is symmetric about the

z

-axis, so

0 \le \theta \le 2\pi

. The volume element is

dV = r \, dz \, dr \, d\theta

. The correct setup is

\int_{0}^{2\pi} \int_{0}^{1} \int_{r}^{2-r^2} r \, dz \, dr \, d\theta

Question 14

Let $R$ be the region in the $xy$ -plane that is inside the circle $x^2+y^2=2x$ but outside the circle $x^2+y^2=1$ . Which of the following integrals represents the area of $R$ ?

$\int_{-\pi/3}^{\pi/3} \int_{2\cos\theta}^{1} r \, dr \, d\theta$
$\int_{-\pi/2}^{\pi/2} \int_{1}^{2\cos\theta} r \, dr \, d\theta$
$\int_{-\pi/3}^{\pi/3} \int_{1}^{2\cos\theta} r \, dr \, d\theta$ (correct answer)
$\int_{-\pi/3}^{\pi/3} \int_{1}^{2\cos\theta} 1 \, dr \, d\theta$

Explanation: When you encounter a region bounded by circles in polar coordinates, your first step is converting the Cartesian equations to polar form and identifying where the boundaries intersect. The circle

x^2 + y^2 = 2x

can be rewritten by completing the square:

(x-1)^2 + y^2 = 1

. This is a circle centered at

(1,0)

with radius 1. In polar coordinates, substituting

x = r\cos\theta

gives us

r^2 = 2r\cos\theta

, so

r = 2\cos\theta

. The circle

x^2 + y^2 = 1

becomes

r = 1

in polar coordinates. To find the intersection points, set

1 = 2\cos\theta

, giving

\cos\theta = \frac{1}{2}

, so

\theta = \pm\frac{\pi}{3}

. This means the region extends from

\theta = -\frac{\pi}{3}

\theta = \frac{\pi}{3}

. For any fixed angle

\theta

in this range,

r

varies from the inner boundary (

r = 1

) to the outer boundary (

r = 2\cos\theta

). The area element in polar coordinates is

r \, dr \, d\theta

, giving us

\int_{-\pi/3}^{\pi/3} \int_{1}^{2\cos\theta} r \, dr \, d\theta

. Option A has the integration limits reversed for

r

. Option B uses incorrect

\theta

limits—at

\theta = \pm\frac{\pi}{2}

, the outer circle has

r = 0

, which doesn't make sense for our region. Option D uses the wrong area element (missing the

r

). Study tip: Always sketch the region first, find intersection points to determine your limits, and remember that polar area elements include the factor

r

Question 15

Consider the region in the first octant bounded by the coordinate planes and the plane $2x + 3y + z = 6$ . When integrating in the order $dz \, dy \, dx$ , what are the correct bounds?

$\int_0^3 \int_0^{(6-2x)/3} \int_0^{6-2x-3y} dz \, dy \, dx$ (correct answer)
$\int_0^6 \int_0^{6-2x} \int_0^{6-2x-3y} dz \, dy \, dx$
$\int_0^3 \int_0^{6-2x} \int_0^{6-2x-3y} dz \, dy \, dx$
$\int_0^2 \int_0^{(6-2x)/3} \int_0^{6-2x-3y} dz \, dy \, dx$

Explanation: The region is in the first octant, so

x \geq 0

y \geq 0

, and

z \geq 0

. The plane

2x + 3y + z = 6

intersects the coordinate axes at

(3,0,0)

(0,2,0)

, and

(0,0,6)

. For the outermost integral (

dx

x

ranges from

0

3

(where the plane intersects the

x

-axis). For the middle integral (

dy

): Given a fixed

x

y

ranges from

0

to where the plane intersects the

xz

-plane at that

x

value. Setting

z = 0

in the plane equation:

2x + 3y = 6

, so

y = (6-2x)/3

. For the innermost integral (

dz

): Given fixed

x

and

y

z

ranges from

0

z = 6 - 2x - 3y

(from the plane equation). Choice B has incorrect upper bound for

x

and wrong middle bounds. Choice C uses

6-2x

instead of

(6-2x)/3

for the

y

bound. Choice D uses

2

instead of

3

as the upper

x

bound.