Problem 20 - Thomas Calculus

Question

Sketch the region of integration and evaluate the integral. \begin{equation} \int_{0}^{\pi} \int_{0}^{\sin x} y d y d x \end{equation}

Step-by-Step Solution

Answer

The value of the integral is \(\frac{\pi}{4}\).

Step 1: Understanding the Integral

The integral given is \[\int_{0}^{\pi} \int_{0}^{\sin x} y \, dy \, dx\].This is a double integral, which means we first integrate with respect to \(y\), and then with respect to \(x\). We need to visualize the limits of integration to understand the region.

Step 2: Sketching the Region of Integration

The inner integral \(\int_{0}^{\sin x} y \, dy\) tells us that for each fixed \(x\), \(y\) ranges from 0 to \(\sin x\). The outer integral \(\int_{0}^{\pi} \, dx\) tells us that \(x\) ranges from 0 to \(\pi\). The region of integration is above the \(x\)-axis, from 0 to \(\pi\), and below the curve \(y=\sin x\). This forms a region bounded by the \(x\)-axis, the lines \(x=0\) and \(x=\pi\), and the curve \(y=\sin x\).

Step 3: Evaluate Inner Integral

Perform the integration with respect to \(y\), keeping \(x\) as a constant:\[\int_{0}^{\sin x} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{\sin x} = \frac{(\sin x)^2}{2} - \frac{0^2}{2} = \frac{(\sin x)^2}{2}.\]

Step 4: Evaluate Outer Integral

Now, substitute the result of the inner integral back into the outer integral:\[\int_{0}^{\pi} \frac{(\sin x)^2}{2} \, dx.\]This is equivalent to\[\frac{1}{2} \int_{0}^{\pi} (\sin x)^2 \, dx.\]

Step 5: Applying Power Reduction Formula

Use the power reduction formula for \((\sin x)^2\):\[(\sin x)^2 = \frac{1 - \cos(2x)}{2}.\]Substitute into the integral:\[\frac{1}{2} \int_{0}^{\pi} \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{4} \int_{0}^{\pi} (1 - \cos(2x)) \, dx.\]

Step 6: Complete the Integration

Now integrate each term:\[\frac{1}{4} \left( \int_{0}^{\pi} 1 \, dx - \int_{0}^{\pi} \cos(2x) \, dx \right).\]Integrate individually:- \(\int_{0}^{\pi} 1 \, dx = [x]_{0}^{\pi} = \pi\)- \(\int_{0}^{\pi} \cos(2x) \, dx = \left[ \frac{\sin(2x)}{2} \right]_{0}^{\pi} = 0\) (since \(\sin(2\pi) = \sin(0) = 0\))Substitute back to get:\[\frac{1}{4} (\pi - 0) = \frac{\pi}{4}.\]

Step 7: Conclusion

The value of the double integral \(\int_{0}^{\pi} \int_{0}^{\sin x} y \, dy \, dx\) is \(\frac{\pi}{4}\).

Key Concepts

Region of Integration

In order to solve a double integral, it is essential to understand the region of integration. This region is essentially the area on the coordinate plane over which the entire integration will be performed. For our given integral, \(\int_{0}^{\pi} \int_{0}^{\sin x} y \, dy \, dx\), we note the following:

• The outer integral \(\int_{0}^{\pi} \, dx\) specifies that \(x\) ranges from 0 to \(\pi\).
• The inner integral \(\int_{0}^{\sin x} y \, dy\) implies that for each fixed \(x\), \(y\) ranges from 0 up to the value of \(\sin x\).

Therefore, the region of integration is bounded by the following:

• The \(x\)-axis (\(y = 0\)),
• The vertical lines \(x = 0\) and \(x = \pi\),
• The curve \(y = \sin x\).

Visualizing this, we are integrating over a region that lies above the \(x\)-axis but below the wave-like curve formed by \(y = \sin x\) from \(x = 0\) to \(x = \pi\). This forms a distinct wavy shape pieced out under the sine curve, setting the stage for our integration.

Inner Integral

When dealing with double integrals, it's important to break down the problem into manageable parts, beginning with the inner integral. This part of the integral involves integrating with respect to \(y\) first while keeping \(x\) constant. In our example, this is given by \(\int_{0}^{\sin x} y \, dy\).

• Integrating \(y\) with respect to itself is straightforward. The integral of \(y\) is \(\frac{y^2}{2}\).
• We then evaluate this at the limits of 0 and \(\sin x\): \(\left[ \frac{y^2}{2} \right]_{0}^{\sin x} = \frac{(\sin x)^2}{2}\).

This result \(\frac{(\sin x)^2}{2}\) becomes the expression that we will substitute into the outer integral. The goal of solving the inner integral first is to reduce the double integration to a single-variable problem, making the next integration step simpler.

Outer Integral

After we solve the inner integral, we proceed to the outer integral. This is performed over the variable \(x\) and makes use of the result from the inner integral. The expression we integrate is now \(\int_{0}^{\pi} \frac{(\sin x)^2}{2} \, dx\), which simplifies to \(\frac{1}{2} \int_{0}^{\pi} (\sin x)^2 \, dx\).

• This evaluation tackles everything in terms of the variable \(x\), integrating the expression over the interval from 0 to \(\pi\).
• The result portrays the aggregate area formed over the defined region of integration when projected on the \(x\)-axis.

Completing this outer integral adds up small slices defined by our previous calculations, culminating in the total value of the double integration. It seamlessly ties both the limits of \(x\) and the previously calculated contribution by \((\sin x)^2\).

Power Reduction Formula

Sometimes, the functions involved in our integral cannot be integrated directly or easily. In such cases, we seek ways to simplify our work. The power reduction formula is one essential tool in calculus that facilitates such simplifications.
In our problem, we encounter \((\sin x)^2\), which can be cumbersome. The power reduction formula offers a handy transformation:

• The formula states that \((\sin x)^2 = \frac{1 - \cos(2x)}{2}\).
• We substitute this into our integral: \(\frac{1}{2} \int_{0}^{\pi} \frac{1 - \cos(2x)}{2} \, dx\).

This substitution reduces our original expression into terms that are easier to integrate directly.
Ultimately, the simplification via power reduction formulae seamlessly aids in executing the integral step-by-step, ensuring an efficient path to the solution.