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📝 Change of Variables in Triple Integrals (14 MCQs)

📖 From Calculus • 15. Multiple Integrals Calculus • 14 questions available

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What is Change of Variables in Triple Integrals?

Definition:
$\iiint_E f(x,y,z) \, dV = \iiint_G f(x(u,v,w), y(u,v,w), z(u,v,w)) \left| \frac{\partial(x,y,z)}{\partial(u,v,w)} \right| \, du \, dv \, dw$ .

Example:
Converting to cylindrical or spherical coordinates uses this formula with the appropriate Jacobian.

Reason:
It generalizes the change of variables to 3D, enabling integration over complex solids by mapping them to simpler regions in new coordinates.

Easy

Medium

Hard

📝 All Change of Variables in Triple Integrals MCQs

Q1. A transformation is defined by $x=2u,\ y=3v,\ z=4w$ . Without evaluating any integral, which factor must multiply the transformed volume element $du\,dv\,dw$ ?

A.-9

B.-12

C.-24 ✅

D.-48

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The transformation scales lengths independently by factors $2,3,$ and $4$ in the three coordinate directions. Therefore, the volume scaling factor is the absolute value of the determinant of the transformation matrix, $2\cdot3\cdot4=24$ . Hence $dV=24\,du\,dv\,dw$ .

Q2. Why is the absolute value of the Jacobian determinant used when transforming a triple integral?

A.It guarantees that the transformed function is positive

B.It accounts for the local change in volume regardless of orientation ✅

C.It removes all nonlinear terms from the integrand

D.It ensures that the transformation is one-to-one everywhere

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A Jacobian determinant can be negative when a transformation reverses orientation, but physical volume cannot be negative. The absolute value converts the signed local scaling factor into the actual volume-scaling factor. Thus, $dV=|J|\,du\,dv\,dw$ , provided the transformation is valid in the region.

Q3. A region is bounded by $x^2+y^2\leq4$ and $0\leq z\leq5$ . Which change of variables most naturally simplifies the description of this region?

x=u+v,\ y=u-v,\ z=w

x=u^2,\ y=v^2,\ z=w^2

x=r\cos\theta,\ y=r\sin\theta,\ z=z

✅

x=r\sin\theta,\ y=r\cos\theta,\ z=z^2

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The circular condition $x^2+y^2\leq4$ becomes $r^2\leq4$ , or $0\leq r\leq2$ , under polar coordinates in the $xy$ -plane. The vertical coordinate remains unchanged. This produces simple bounds and introduces the volume factor $r$ , making the integral substantially easier to formulate.

Q4. Suppose $x=u+v,\ y=u-v,\ z=w$ . A student claims that the volume element remains $du\,dv\,dw$ because each original variable is a linear combination of the new variables. What is the best evaluation of this claim?

A.Correct, because linear transformations never change volume

B.Correct, because the transformation has determinant

1

C.Incorrect, because the absolute Jacobian is

2

✅

D.Incorrect, because the absolute Jacobian is

1/2

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The transformation matrix is $\begin{pmatrix}1&1&0\\1&-1&0\\0&0&1\end{pmatrix}$ , whose determinant is $-2$ . Therefore, the volume scaling factor is $|-2|=2$ . The correct differential relation is $dV=2\,du\,dv\,dw$ , so linearity alone does not imply unit volume scaling.

Q5. A solid is described by $x^2+y^2+z^2\leq9$ . Which transformed bounds are most appropriate when spherical coordinates are used with $x=\rho\sin\phi\cos\theta,\ y=\rho\sin\phi\sin\theta,\ z=\rho\cos\phi$ ?

0\leq\rho\leq3,\ 0\leq\phi\leq\pi,\ 0\leq\theta\leq2\pi

✅

0\leq\rho\leq9,\ 0\leq\phi\leq\pi/2,\ 0\leq\theta\leq2\pi

0\leq\rho\leq3,\ 0\leq\phi\leq2\pi,\ 0\leq\theta\leq\pi

0\leq\rho\leq9,\ 0\leq\phi\leq\pi,\ 0\leq\theta\leq\pi

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Since $x^2+y^2+z^2=\rho^2$ , the spherical region becomes $0\leq\rho\leq3$ . To cover the entire sphere, the polar angle must range from $0$ to $\pi$ , while the azimuthal angle ranges from $0$ to $2\pi$ . These bounds cover the solid exactly once.

Q6. A manufacturing model describes a solid by $x^2+y^2\leq16$ , $0\leq z\leq x^2+y^2$ . Which coordinate substitution gives the most direct description of the region?

A.Cartesian coordinates, because all variables already appear explicitly

B.Cylindrical coordinates, because both boundaries depend on

x^2+y^2

✅

C.Spherical coordinates, because the region contains a quadratic term

D.The substitution

x=u^2,\ y=v^2,\ z=w

, because it eliminates all squares

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The repeated expression $x^2+y^2$ signals rotational symmetry around the $z$ -axis. With $x=r\cos\theta$ and $y=r\sin\theta$ , the region becomes $0\leq r\leq4$ , $0\leq\theta\leq2\pi$ , and $0\leq z\leq r^2$ . This avoids unnecessarily complicated Cartesian bounds.

Q7. Consider $x=uv,\ y=u/v,\ z=w$ , where $u>0$ and $v>0$ . If a region is naturally described by bounds on $x/y$ and $xy$ , why might this transformation be useful?

A.Because it can convert products and ratios into simple expressions involving

u

and

v

✅

B.Because its Jacobian is always equal to

1

C.Because it automatically converts every boundary into a plane

D.Because it eliminates the need to transform the integrand

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Under this substitution, $xy=(uv)(u/v)=u^2$ , while $x/y=(uv)/(u/v)=v^2$ . Thus, boundaries involving $xy$ and $x/y$ can become simple bounds in $u$ and $v$ . The Jacobian still must be calculated, so the transformation simplifies geometry rather than eliminating all integration work.

Q8. A student changes variables using $x=2u,\ y=2v,\ z=3w$ , but writes $dV=12\,du\,dv\,dw$ . Which statement correctly identifies the mistake?

A.The factor should be

6

because only two variables were scaled

B.The factor should be

7

because the scaling factors are added

C.The factor should be

12

only when the transformation reverses orientation

D.The factor should be

12

, so there is no mistake ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The determinant of the diagonal transformation matrix is $2\cdot2\cdot3=12$ . Therefore, $dV=12\,du\,dv\,dw$ . The student has correctly multiplied the three independent scaling factors. A common misconception is to add scaling factors instead of multiplying them.

Q9. A region has the shape of a cone $z=\sqrt{x^2+y^2}$ above the $xy$ -plane and is cut off by $z=4$ . Which spherical-coordinate relationship best describes the conical boundary?

\phi=\pi/4

✅

\rho=4

\theta=\pi/4

\rho\cos\phi=4

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: In spherical coordinates, $z=\rho\cos\phi$ and $\sqrt{x^2+y^2}=\rho\sin\phi$ . The cone equation becomes $\rho\cos\phi=\rho\sin\phi$ , so $\tan\phi=1$ , giving $\phi=\pi/4$ . The plane $z=4$ then becomes $\rho\cos\phi=4$ , supplying the radial boundary.

Q10. A graph shows a solid symmetric about the $z$ -axis whose horizontal cross-sections are circles centered at the origin. The radius increases with height according to $r=z$ , for $0\leq z\leq2$ . Which bounds best model the solid in cylindrical coordinates?

0\leq r\leq2,\ 0\leq\theta\leq2\pi,\ 0\leq z\leq r

0\leq r\leq z,\ 0\leq\theta\leq2\pi,\ 0\leq z\leq2

✅

0\leq r\leq z,\ 0\leq\theta\leq\pi,\ 0\leq z\leq2

0\leq r\leq2z,\ 0\leq\theta\leq2\pi,\ 0\leq z\leq2

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The graph indicates rotational symmetry, so cylindrical coordinates are natural. Since the radius satisfies $r=z$ , each horizontal slice has $0\leq r\leq z$ . Full rotation requires $0\leq\theta\leq2\pi$ , and the height restriction gives $0\leq z\leq2$ .

Q11. A triple integral over a spherical region contains the integrand $x^2+y^2+z^2$ . A student says the substitution is unnecessary because the expression already resembles a square. Which response is most mathematically appropriate?

A.The student is correct because the integrand cannot be simplified further

B.The substitution can still be useful because the integrand becomes

\rho^2

and the volume element becomes

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

✅

C.The substitution is invalid because spherical coordinates cannot represent quadratic expressions

D.The substitution changes the integrand to

\rho\sin\phi

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Although the original expression is recognizable, spherical coordinates simultaneously simplify both the geometry and the integrand. Since $x^2+y^2+z^2=\rho^2$ , and $dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ , the transformed integral becomes a product of simple factors with straightforward bounds.

Q12. Two methods are proposed for evaluating a volume bounded by $x^2+y^2\leq4$ , $z\geq0$ , and $z\leq4-x^2-y^2$ . Method I uses Cartesian coordinates, while Method II uses cylindrical coordinates. Which comparison is most accurate?

A.Method I necessarily gives shorter bounds because Cartesian coordinates are fundamental

B.Method II is usually more efficient because radial symmetry converts

x^2+y^2

into

r^2

✅

C.Both methods must have identical differential elements

D.Method II is invalid because cylindrical coordinates cannot describe surfaces involving

z

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The region has circular symmetry around the $z$ -axis. In cylindrical coordinates, the conditions become $0\leq r\leq2$ , $0\leq\theta\leq2\pi$ , and $0\leq z\leq4-r^2$ , with $dV=r\,dz\,dr\,d\theta$ . Cartesian coordinates require a more cumbersome description involving $x^2+y^2$ .

Q13. Let $x=u+v,\ y=u-v,\ z=2w$ . A student computes the Jacobian as $1\cdot(-1)\cdot2=-2$ and concludes that the transformed volume is negative. What should the student conclude instead?

A.The transformation is impossible because a Jacobian cannot be negative

B.The volume factor is

2

, because the absolute value of the Jacobian is required ✅

C.The volume factor is

-2

, because orientation determines physical volume

D.The volume factor is

1/2

, because the determinant must be inverted

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The determinant is indeed $-2$ , so the student's determinant calculation is not the problem. A negative determinant indicates orientation reversal, not negative physical volume. The change in volume is governed by the absolute value, $|J|=2$ , giving $dV=2\,du\,dv\,dw$ .

Q14. For a transformation whose Jacobian determinant is $J(u,v,w)=u^2+v^2+w^2$ , a region excludes the origin. Which conclusion is most important when setting up a transformed triple integral?

A.The Jacobian can be ignored because it is not constant

B.The transformation is locally volume-preserving everywhere

C.The volume element must include

(u^2+v^2+w^2)\,du\,dv\,dw

, provided the transformation is one-to-one on the region ✅

D.The volume element must include

1/(u^2+v^2+w^2)\,du\,dv\,dw

regardless of the transformation

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The Jacobian gives the local scaling of volume under the transformation. Since $u^2+v^2+w^2>0$ away from the origin, its absolute value equals the expression itself. Therefore, $dV=(u^2+v^2+w^2)\,du\,dv\,dw$ , assuming the transformation is valid and one-to-one over the region.

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