Volume Of Revolution & Distance Traveled Problem Solved

Hey guys! Today, we're diving deep into the fascinating world of calculus, tackling two super interesting problems. First, we'll figure out the volume of a solid formed by rotating a shape around the x-axis. Then, we'll use calculus to determine the distance an object travels given its velocity. Get ready to put on your math hats – it's gonna be an exciting ride!

1. Finding the Volume of Revolution

So, our first challenge is to calculate the volume of the solid created when we rotate a specific 2D shape around the x-axis. This shape, called a curvilinear trapezoid, is bounded by these lines:

• y = x² - 1
• x = 1
• x = 2
• y = 0

To visualize this, imagine the curve y = x² - 1 nestled between the vertical lines x = 1 and x = 2, and sitting above the x-axis (y = 0). When we spin this little slice of the graph around the x-axis, it forms a 3D shape – a solid of revolution. Our mission, should we choose to accept it, is to figure out the volume of this spunky solid.

The Disk Method: Our Weapon of Choice

The key to solving this lies in a method called the disk method. Think of it like this: we're going to slice our 3D solid into a bunch of infinitely thin disks, all stacked next to each other. Each disk has a tiny thickness (dx) and a radius that corresponds to the function's y-value at a given x. The volume of each individual disk is simply the area of a circle (πr²) multiplied by its thickness (dx).

So, the volume (dV) of a single disk can be expressed as:

dV = π [f(x)]² dx

Where f(x) represents the function defining our curve (in this case, y = x² - 1).

Setting Up the Integral: The Blueprint for Calculation

To get the total volume of the solid, we need to sum up the volumes of all these infinitely thin disks. This, my friends, is where the magic of integration comes in! We'll integrate the disk volume formula over the interval defined by our limits of integration (the x-values that bound our shape).

In our case, the limits of integration are x = 1 and x = 2. So, the integral representing the total volume (V) looks like this:

V = ∫[from 1 to 2] π [x² - 1]² dx

Cracking the Code: Solving the Integral

Alright, let's get our hands dirty and solve this integral! First, we need to expand the expression inside the integral:

(x² - 1)² = x⁴ - 2x² + 1

Now, our integral looks like this:

V = π ∫[from 1 to 2] (x⁴ - 2x² + 1) dx

Next, we find the antiderivative of each term:

∫ x⁴ dx = (x⁵)/5 ∫ -2x² dx = (-2x³)/3 ∫ 1 dx = x

Putting it all together, the antiderivative of our integrand is:

(x⁵)/5 - (2x³)/3 + x

Now, we evaluate this antiderivative at our limits of integration (2 and 1) and subtract:

[((2)⁵)/5 - (2(2)³)/3 + 2] - [((1)⁵)/5 - (2(1)³)/3 + 1]

Let's break it down step by step:

• Evaluate at x=2:
  • (2⁵)/5 = 32/5
  • (2(2)³)/3 = 16/3
  • 32/5 - 16/3 + 2 = (96 - 80 + 30)/15 = 46/15
• Evaluate at x=1:
  • (1⁵)/5 = 1/5
  • (2(1)³)/3 = 2/3
  • 1/5 - 2/3 + 1 = (3 - 10 + 15)/15 = 8/15
• Subtract: 46/15 - 8/15 = 38/15

Finally, we multiply by π:

V = π (38/15)

So, the volume of the solid of revolution is (38π)/15 cubic units. Woohoo! We conquered the volume!

2. Decoding Motion: Finding Distance Traveled

Now, let's shift gears and tackle a different kind of problem. We're given the velocity function of an object moving in a straight line:

v(t) = -6t² + 2t

Our mission, should we choose to accept it (again!), is to determine the total distance the object travels between time t = 1 and t = 3. This isn't just about displacement (the net change in position); we want to know the total ground the object covered, even if it changed direction.

The Integral Connection: Linking Velocity and Distance

The key concept here is the relationship between velocity and distance. Velocity is the rate of change of position, and distance traveled is the accumulation of this rate over time. Mathematically, this means we need to integrate the absolute value of the velocity function over the given time interval.

Why the absolute value? Because distance is always positive, even if the object is moving in the negative direction. The integral of the regular velocity function would give us the displacement, which could be negative if the object moves backward more than it moves forward.

So, the formula for total distance (D) is:

D = ∫[from 1 to 3] |v(t)| dt = ∫[from 1 to 3] |-6t² + 2t| dt

Handling the Absolute Value: Finding Turning Points

The tricky part is dealing with the absolute value. To do this properly, we need to figure out if the velocity changes sign (meaning the object changes direction) within our time interval. This happens when the velocity is equal to zero.

Let's find the times when v(t) = 0:

-6t² + 2t = 0

We can factor out a 2t:

2t(-3t + 1) = 0

This gives us two solutions:

• t = 0
• t = 1/3

Only t = 1/3 falls within our interval of interest (t = 1 to t = 3). This means the object changes direction at t = 1/3. But since 1/3 is not within our interval [1, 3], the velocity does not change sign within our interval. So we do not need to consider separate intervals.

Setting up the Integral (Again!): Accounting for Direction

Since the velocity might be negative over the interval, we need to consider the sign of v(t) = -6t² + 2t on the interval [1, 3]. Let's pick a test point, say t = 2:

v(2) = -6(2)² + 2(2) = -24 + 4 = -20

Since v(2) is negative, the velocity is negative on the interval [1, 3]. This means we need to take the negative of the velocity function inside the integral to account for the absolute value:

D = ∫[from 1 to 3] |-6t² + 2t| dt = ∫[from 1 to 3] -(-6t² + 2t) dt = ∫[from 1 to 3] (6t² - 2t) dt

Solving the Integral: The Final Stretch

Now we're in familiar territory! Let's find the antiderivative of 6t² - 2t:

∫ 6t² dt = 2t³ ∫ -2t dt = -t²

So, the antiderivative is:

2t³ - t²

Evaluate at the limits of integration:

[2(3)³ - (3)²] - [2(1)³ - (1)²]

Let's break it down:

• Evaluate at t=3:
  • 2(3)³ - (3)² = 54 - 9 = 45
• Evaluate at t=1:
  • 2(1)³ - (1)² = 2 - 1 = 1
• Subtract: 45 - 1 = 44

Therefore, the total distance traveled by the object from t = 1 to t = 3 is 44 units. We did it!

Wrapping Up: Calculus Conquerors!

Guys, we've tackled some seriously cool calculus problems today! We learned how to find the volume of a solid of revolution using the disk method and how to calculate the total distance traveled by an object given its velocity function. These are powerful tools that can be applied to a wide range of real-world scenarios. So, keep practicing, keep exploring, and keep conquering those mathematical challenges!