C656 MATH 3311
Calculus III
COMPLETE FA REVIEW
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,1. Question:
What condition must a vector field F satisfy to be conservative?
a) \( \nabla \cdot \mathbf{F} = 0 \)
b) \( \nabla \times \mathbf{F} = 0 \)
c) \( \nabla \mathbf{F} = 0 \)
d) \( \int_{C} \mathbf{F} \cdot d\mathbf{r} = 0 \)
Answer: b) \( \nabla \times \mathbf{F} = 0 \)
Rationale: A vector field is conservative if and only if its curl is
zero.
2. Question:
Evaluate the integral \( \int_{0}^{1} \int_{0}^{x} xy \, dy \, dx \).
a) \( \frac{1}{8} \)
b) \( \frac{1}{4} \)
c) \( \frac{1}{6} \)
d) \( \frac{1}{2} \)
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, Answer: b) \( \frac{1}{4} \)
Rationale: First, integrate with respect to \(y\), giving \(
\int_{0}^{1} \left[ \frac{1}{2}xy^2 \right]_0^x dx \), resulting in \(
\frac{1}{2} \int_{0}^{1} x^3 \, dx = \frac{1}{4} \).
3. Question:
Which of the following integrals represents the surface area of
the paraboloid \( z = 4 - x^2 - y^2 \) for \( z \geq 0 \)?
a) \( \iint_{D} 1 \, dA \)
b) \( \iint_{D} \sqrt{1 + (\nabla z)^2} \, dA \)
c) \( \iint_{D} \sqrt{1 + (2x)^2 + (2y)^2} \, dA \)
d) \( \iint_{D} \sqrt{16 - x^2 - y^2} \, dA \)
Answer: c) \( \iint_{D} \sqrt{1 + (2x)^2 + (2y)^2} \, dA \)
Rationale: The surface area integral formula over a region \(D\)
requires the integrand \( \sqrt{1 + (\frac{\partial z}{\partial x})^2 +
(\frac{\partial z}{\partial y})^2} \).
4. Question:
© 2024
Calculus III
COMPLETE FA REVIEW
© 2024
,1. Question:
What condition must a vector field F satisfy to be conservative?
a) \( \nabla \cdot \mathbf{F} = 0 \)
b) \( \nabla \times \mathbf{F} = 0 \)
c) \( \nabla \mathbf{F} = 0 \)
d) \( \int_{C} \mathbf{F} \cdot d\mathbf{r} = 0 \)
Answer: b) \( \nabla \times \mathbf{F} = 0 \)
Rationale: A vector field is conservative if and only if its curl is
zero.
2. Question:
Evaluate the integral \( \int_{0}^{1} \int_{0}^{x} xy \, dy \, dx \).
a) \( \frac{1}{8} \)
b) \( \frac{1}{4} \)
c) \( \frac{1}{6} \)
d) \( \frac{1}{2} \)
© 2024
, Answer: b) \( \frac{1}{4} \)
Rationale: First, integrate with respect to \(y\), giving \(
\int_{0}^{1} \left[ \frac{1}{2}xy^2 \right]_0^x dx \), resulting in \(
\frac{1}{2} \int_{0}^{1} x^3 \, dx = \frac{1}{4} \).
3. Question:
Which of the following integrals represents the surface area of
the paraboloid \( z = 4 - x^2 - y^2 \) for \( z \geq 0 \)?
a) \( \iint_{D} 1 \, dA \)
b) \( \iint_{D} \sqrt{1 + (\nabla z)^2} \, dA \)
c) \( \iint_{D} \sqrt{1 + (2x)^2 + (2y)^2} \, dA \)
d) \( \iint_{D} \sqrt{16 - x^2 - y^2} \, dA \)
Answer: c) \( \iint_{D} \sqrt{1 + (2x)^2 + (2y)^2} \, dA \)
Rationale: The surface area integral formula over a region \(D\)
requires the integrand \( \sqrt{1 + (\frac{\partial z}{\partial x})^2 +
(\frac{\partial z}{\partial y})^2} \).
4. Question:
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