Problem 16
Question
Evaluate the line integral. \(\int_{C} 3 y^{2} d y,\) where \(C\) is the portion of \(y=x^{2}\) from (2,4) to (0,0)
Step-by-Step Solution
Verified Answer
The value of the line integral is 19.2.
1Step 1: Parameterize the Curve
The given curve is \(y = x^{2}\) and we are moving from point (2,4) to point (0,0). We need to parameterize this curve. To do this, we usually set one of our variables equal to the parameter. Let's set \(x = t\). Because \(y = x^{2}\), we can say that \(y = t^{2}\). We choose the limits for \(t\) based on the given points. Since \(x\) ranges from 2 to 0, \(t\) will also range from 2 to 0.
2Step 2: Substitute the Parameterized Function into the Integral
Substitute \(y = t^{2}\) into the integral. This makes the integral become \(\int_{2}^{0} 3 t^{4} dt\), where our new limits are from 2 to 0.
3Step 3: Evaluate the Integral
We now evaluate the integral. The antiderivative of \(3 t^{4}\) is \(\frac{3}{5} t^{5}\). Evaluating this definite integral from 2 to 0 gives us \(\frac{3}{5} * 0 - \frac{3}{5} * 32 = - 19.2\). However, it's a convention that we do not consider the direction of the path when computing line integrals over scalar fields, so we'll change the sign. The result is 19.2.
Key Concepts
Parameterizing the CurveEvaluating the IntegralFinding the Antiderivative
Parameterizing the Curve
Parameterizing a curve is a way of expressing the variables of a curve in terms of a parameter, usually denoted by a letter like \(t\). This transformation makes it easier to evaluate integrals along curves. In our exercise, we are working with a curve defined by the equation \(y = x^2\), starting at the point (2,4) and moving to (0,0).
To parameterize this curve, we set \(x=t\). Since \(y=x^2\), we substitute \(x\) with \(t\) to get \(y=t^2\). Now, both \(x\) and \(y\) are expressed in terms of \(t\).
To parameterize this curve, we set \(x=t\). Since \(y=x^2\), we substitute \(x\) with \(t\) to get \(y=t^2\). Now, both \(x\) and \(y\) are expressed in terms of \(t\).
- For \(t=2\), we have the point (2, 4).
- For \(t=0\), we reach the endpoint (0, 0).
Evaluating the Integral
Once a curve is parameterized, the next step is substituting this parameterization into the integral expression. In our case, we began with the line integral \(\int_{C} 3y^2 \, dy\).
By substituting \(y=t^2\) into the integral, we change the original line integral into an ordinary integral with respect to \(t\). So, it becomes \(\int_{2}^{0} 3t^4 \, dt\). It's important to remember that when setting the bounds from \(2\) to \(0\), we account for the direction of traversal along the curve. However, scalar line integrals are independent of the path direction when evaluating.
This substitution simplifies solving the problem by turning the line integral into a standard integral format, which we can then evaluate using basic integral calculus techniques.
By substituting \(y=t^2\) into the integral, we change the original line integral into an ordinary integral with respect to \(t\). So, it becomes \(\int_{2}^{0} 3t^4 \, dt\). It's important to remember that when setting the bounds from \(2\) to \(0\), we account for the direction of traversal along the curve. However, scalar line integrals are independent of the path direction when evaluating.
This substitution simplifies solving the problem by turning the line integral into a standard integral format, which we can then evaluate using basic integral calculus techniques.
Finding the Antiderivative
Evaluating an integral typically involves finding its antiderivative, which is the reverse process of differentiation. For our integral, \(\int_{2}^{0} 3t^4 \, dt\), we need to find the antiderivative of \(3t^4\).
The antiderivative of \(t^n\) is \(\frac{t^{n+1}}{n+1}\). So, the antiderivative of \(3t^4\) is \(\frac{3}{5}t^5\). To evaluate the definite integral, plug the bounds into the antiderivative:
- At \(t=0\), the antiderivative is \(\frac{3}{5} imes 0^5 = 0\).
- At \(t=2\), the antiderivative is \(\frac{3}{5} imes 2^5 = \frac{3}{5} imes 32 = 19.2\).
The definite integral equals the difference between the two values, taking into account any necessary sign changes that arise from the convention of path direction. Thus, we get \(19.2\). Understanding the antiderivative is crucial in converting a given function into its integral form, which then allows us to evaluate the area under the curve or along a path.
The antiderivative of \(t^n\) is \(\frac{t^{n+1}}{n+1}\). So, the antiderivative of \(3t^4\) is \(\frac{3}{5}t^5\). To evaluate the definite integral, plug the bounds into the antiderivative:
- At \(t=0\), the antiderivative is \(\frac{3}{5} imes 0^5 = 0\).
- At \(t=2\), the antiderivative is \(\frac{3}{5} imes 2^5 = \frac{3}{5} imes 32 = 19.2\).
The definite integral equals the difference between the two values, taking into account any necessary sign changes that arise from the convention of path direction. Thus, we get \(19.2\). Understanding the antiderivative is crucial in converting a given function into its integral form, which then allows us to evaluate the area under the curve or along a path.
Other exercises in this chapter
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