Problem 337 - Calculus Volume 3

Question

Find the area of the region. $$ x=t^{2}, \quad y=\ln (t), \quad 0 \leq t \leq e $$

Step-by-Step Solution

Answer

The area is \( \frac{e^2}{2} \).

Step 1: Understand the Parametric Equations

The problem provides a set of parametric equations: \( x = t^2 \) and \( y = \ln(t) \), where \( t \) ranges from \( 0 \) to \( e \). We are tasked with finding the area of the region defined by these equations within the given range of \( t \).

Step 2: Set Up the Integral for Area

To find the area of a region defined by parametric equations, we use the formula for the area \( A \) between the parametric curve and the x-axis: \[ A = \int_{t=a}^{t=b} y \frac{dx}{dt} \, dt \] Substitute \( y = \ln(t) \) and differentiate \( x = t^2 \) with respect to \( t \): \[ \frac{dx}{dt} = 2t \] This gives us:\[ A = \int_{0}^{e} \ln(t) \cdot 2t \, dt \]

Step 3: Evaluate the Integral

The integral to solve is \( \int_{0}^{e} 2t \ln(t) \, dt \). Use integration by parts, where \( u = \ln(t) \) and \( dv = 2t \, dt \). Then, \( du = \frac{1}{t} \dt \) and \( v = t^2 \).The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Apply it here:\[\int 2t \ln(t) \, dt = [ \ln(t) \cdot t^2 ]_{0}^{e} - \int_0^e t^2 \cdot \frac{1}{t} \, dt\]Simplify and evaluate the new integral:\[ [ t^2 \ln(t) ]_{0}^{e} - \int_0^e t \, dt \]

Step 4: Compute Each Term of the Integral

Calculate each term:- \([ t^2 \ln(t) ]_0^e = e^2 \ln(e) - 0^2 \ln(0) = e^2 \cdot 1 = e^2 \) (as \( \ln(e) = 1 \) and \( 0^2 \cdot \ln(0) \) approaches zero)- The integral \( \int_0^e t \, dt \) is computed as \([ \frac{t^2}{2} ]_0^e = \frac{e^2}{2} - 0 = \frac{e^2}{2} \)Combine these results to find the area:\[ e^2 - \frac{e^2}{2} = \frac{e^2}{2} \]

Step 5: State the Final Area

The area of the region bounded by the parametric equations \( x = t^2 \), \( y = \ln(t) \) from \( t = 0 \) to \( t = e \) is \( \frac{e^2}{2} \). This is the area under the curve \( y \) as defined by the parametric equations.

Key Concepts

Integration by Parts

Integration by parts is a valuable technique used to solve integrals where the standard antiderivative methods do not easily apply. This method is particularly useful when dealing with products of functions. The integration by parts formula is given by:\[ \int u \, dv = uv - \int v \, du \]Here, we assign parts of the original integral to functions \( u \) and \( dv \). After differentiating \( u \) to find \( du \) and integrating \( dv \) to find \( v \), substitute them into the formula.

In this exercise, we apply integration by parts to evaluate the integral \( \int 2t \ln(t) \, dt \). By choosing \( u = \ln(t) \) and \( dv = 2t \, dt \), we differentiate and integrate respectively to find \( du = \frac{1}{t} \, dt \) and \( v = t^2 \). Substituting back into the integration by parts formula simplifies the computation of the integral. This leads us closer to finding the area under the parametric curve.

Area Under the Curve

The concept of finding the area under the curve is crucial in calculus, representing a fundamental way to measure the size of regions under a graphed function. For non-parametric functions, this typically involves integrating with respect to one variable over a specified interval. When dealing with curves defined by parametric equations, the process becomes a tad more complex.

For a curve defined parametrically, such as \( x = t^2 \) and \( y = \ln(t) \), the formula to find the area \( A \) between the curve and the x-axis is given by:\[ A = \int_{t=a}^{t=b} y \frac{dx}{dt} \, dt \]This requires us to express \( y \) in terms of \( t \) and to compute \( \frac{dx}{dt} \), which involves differentiating \( x \) with respect to \( t \). Over the range from \( t = 0 \) to \( t = e \), we substitute \( y = \ln(t) \) and \( \frac{dx}{dt} = 2t \) into the integral. Solving this integral gives the area under the curve specified by the parametric equations within those bounds.

Parametric Area Calculation

Parametric area calculation involves integrating products involving derivatives when the path of the curve is given in parametric form. Unlike the usual \(y\) as a function of \(x\), parametric equations use a third variable, often \(t\), to express both \(x\) and \(y\).

To find the area under the curve described by parametric equations, we use the formula:\[ A = \int_{t=a}^{t=b} y \frac{dx}{dt} \, dt \]This approach leverages the derivative \( \frac{dx}{dt} \) to account for changes along the x-axis as \( t \) progresses. The exercise illustrates this method by finding the area of a region with equations \( x = t^2 \) and \( y = \ln(t) \). First, differentiate \( x \) to get \( \frac{dx}{dt} = 2t \), and then place both \( y \) and \( \frac{dx}{dt} \) into the integral for calculation over the interval from \( t = 0 \) to \( t = e \).
The final result, \( \frac{e^2}{2} \), reveals the area of the region defined by the parametric equations, showcasing the effectiveness of this method in accounting for areas under curves in parametric form.