Calculus II Review: Arc Length, Surface Area, Physical Applications, and Exponential Models

Q1. Find the arc length of the curve y = -8x - 3 on [−2, 6].

Background

Topic: Arc Length of a Curve

This question tests your ability to compute the arc length of a function over a given interval using calculus.

Key Terms and Formulas

• Arc Length Formula: For a curve $y = f(x)$ from $x = a$ to $x = b$, the arc length $L$ is given by:

$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Step-by-Step Guidance

1. Find $\frac{dy}{dx}$ for $y = -8x - 3$.

2. Square $\frac{dy}{dx}$ and add 1 inside the square root.

3. Set up the definite integral for arc length from $x = -2$ to $x = 6$ using the formula above.

4. Simplify the integrand as much as possible before integrating.

Try solving on your own before revealing the answer!

Q2. Find the arc length of $y = \sqrt{x} + 2$ on [1, 2].

Background

Topic: Arc Length of a Curve

This question asks you to apply the arc length formula to a function involving a square root.

Key Terms and Formulas

$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Step-by-Step Guidance

1. Compute $\frac{dy}{dx}$ for $y = \sqrt{x} + 2$.

2. Square $\frac{dy}{dx}$ and add 1.

3. Set up the definite integral for arc length from $x = 1$ to $x = 2$.

4. Simplify the integrand as much as possible.

Try solving on your own before revealing the answer!

Q3. Find the arc length of $y = x^{3/2} - x^{1/2}$ on [1, 9].

Background

Topic: Arc Length of a Curve

This problem involves powers of $x$ and requires careful differentiation and algebraic manipulation.

Key Terms and Formulas

$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Step-by-Step Guidance

1. Find $\frac{dy}{dx}$ for $y = x^{3/2} - x^{1/2}$.

2. Square $\frac{dy}{dx}$ and add 1.

3. Set up the definite integral for arc length from $x = 1$ to $x = 9$.

4. Simplify the integrand as much as possible before integrating.

Try solving on your own before revealing the answer!

Q4. Golden Gate cables: Approximate the length of the cables that stretch between the tops of the two towers. (Setup integral only)

Background

Topic: Arc Length of a Parabola (Applied Problem)

This question models the cable of a suspension bridge as a parabola and asks you to set up (not evaluate) the integral for its length.

Key Terms and Formulas

• Given: $y = 0.00037x^2$, $|x| \leq 640$ (so $x$ runs from $-640$ to $640$)

• Arc Length Formula:

$L = \int_{-640}^{640} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Step-by-Step Guidance

1. Find $\frac{dy}{dx}$ for $y = 0.00037x^2$.

2. Square $\frac{dy}{dx}$ and add 1 inside the square root.

3. Write the definite integral for arc length from $x = -640$ to $x = 640$.

4. Do not evaluate the integral; just set it up as requested.

Try setting up the integral before revealing the answer!

Q5. Cosine vs. parabola: Which curve has the greater length on the interval [−1, 1], $y = 1 - x^2$, or $y = \cos \pi x$? (Setup integral only)

Background

Topic: Comparing Arc Lengths of Different Curves

This question asks you to set up (not evaluate) the arc length integrals for two different curves over the same interval.

Key Terms and Formulas

$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Step-by-Step Guidance

1. For each curve, compute $\frac{dy}{dx}$.

2. Square $\frac{dy}{dx}$ and add 1 for each curve.

3. Set up the definite integral for arc length from $x = -1$ to $x = 1$ for both curves.

4. Do not evaluate the integrals; just write them out for comparison.

Try setting up both integrals before revealing the answer!

Q6. Find the area of the surface generated when $y = 3x + 4$, $0 \leq x \leq 6$, is revolved about the x-axis.

Background

Topic: Surface Area of Revolution

This question tests your ability to set up and compute the surface area generated by revolving a curve about the x-axis.

Key Terms and Formulas

• Surface Area Formula (about x-axis):

$S = 2\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Step-by-Step Guidance

1. Find $\frac{dy}{dx}$ for $y = 3x + 4$.

2. Square $\frac{dy}{dx}$ and add 1 inside the square root.

3. Set up the definite integral for surface area from $x = 0$ to $x = 6$ using the formula above.

4. Simplify the integrand as much as possible before integrating.

Try solving on your own before revealing the answer!

Q7. A 1.5 mm layer of paint is applied to one side of the surface generated when $y = 8x - x^2$ on $1 \leq x \leq 7$ is revolved about the x-axis. Find the approximate volume of paint needed.

Background

Topic: Surface Area and Volume Applications

This problem combines surface area of revolution with a real-world application: finding the volume of a thin layer (paint) covering a surface.

Key Terms and Formulas

• Surface Area Formula (about x-axis):

$S = 2\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

• Volume of paint $\approx$ Surface Area $\times$ thickness (convert thickness to meters!)

Step-by-Step Guidance

1. Find $\frac{dy}{dx}$ for $y = 8x - x^2$.

2. Set up the surface area integral from $x = 1$ to $x = 7$.

3. Multiply the resulting surface area by the thickness (in meters) to get the volume.

4. Do not compute the final value; just set up the expressions.

Try setting up the expressions before revealing the answer!

Q8. Zones of a sphere: Show that the surface area of the resulting zone on the sphere is $2\pi r h$, independent of the location of the cutting planes.

Background

Topic: Surface Area of a Spherical Zone

This question asks you to derive a formula for the surface area of a spherical zone (the region between two parallel planes cutting a sphere).

Key Terms and Formulas

• Surface area of a sphere: $4\pi r^2$

• Zone height: $h$

• Radius: $r$

Step-by-Step Guidance

1. Visualize the zone as a "band" around the sphere of height $h$.

2. Recall that the lateral surface area of a cylinder of radius $r$ and height $h$ is $2\pi r h$.

3. Argue (using calculus or geometry) why the surface area of the spherical zone is the same as that of a cylinder of the same radius and height.

4. Set up the integral or geometric argument, but do not complete the proof.

Try reasoning through the setup before revealing the answer!

Q9. Emptying a conical tank: A water tank is shaped like an inverted cone with height 6 m and base radius 1.5 m. If the tank is full, how much work is required to pump the water to the level of the top of the tank and out of the tank?

Background

Topic: Work Done by Pumping Fluids

This question involves setting up an integral to compute the work required to pump water out of a tank with a conical shape.

Key Terms and Formulas

• Work: $W = \int_a^b F(x) \, dx$

• For fluids: $W = \int_0^H (\text{weight of slice}) \times (\text{distance to move slice}) \, dy$

• Volume of a thin slice: $dV = \pi r^2(y) \, dy$

• Density of water: $\rho = 1000 \text{ kg/m}^3$

• Gravity: $g = 9.8 \text{ m/s}^2$

Step-by-Step Guidance

1. Express the radius $r$ of a horizontal slice at height $y$ in terms of $y$ using similar triangles.

2. Write the volume of a thin slice at height $y$ as $dV = \pi r^2(y) \, dy$.

3. Find the weight of the slice: $\text{weight} = \rho g dV$.

4. Set up the integral for work, where the distance each slice must be lifted is $(6 - y)$.

5. Write the definite integral for $y$ from $0$to $6$.

Try setting up the integral before revealing the answer!

Q10. Emptying a water trough: A water trough has a semicircular cross section with a radius of 0.25 m and a length of 3 m. How much work is required to pump the water out of the trough when it is full?

Background

Topic: Work Done by Pumping Fluids

This problem involves setting up an integral to compute the work required to pump water out of a trough with a semicircular cross-section.

Key Terms and Formulas

• Work: $W = \int_a^b F(x) \, dx$

• For fluids: $W = \int_0^R (\text{weight of slice}) \times (\text{distance to move slice}) \, dy$

• Area of a horizontal slice at height $y$ in a semicircle: $A(y) = 2 \times \text{length} \times \sqrt{r^2 - y^2}$

• Density of water: $\rho = 1000 \text{ kg/m}^3$

• Gravity: $g = 9.8 \text{ m/s}^2$

Step-by-Step Guidance

1. Express the width of the water surface at height $y$ using the semicircle equation.

2. Write the area of a horizontal slice at height $y$.

3. Find the weight of the slice: $\text{weight} = \rho g A(y) \, dy$.

4. Set up the integral for work, where the distance each slice must be lifted is $(0.25 - y)$.

5. Write the definite integral for $y$ from $0$to $0.25$.

Try setting up the integral before revealing the answer!