Question

Let S be the triangular region whose vertices are (0, 0), \((0,\frac{\pi}{2})\) and \((\frac{\pi}{2},0)\). The value of \(\iint\limits_S\sin(x)\cos(y)dx\ dy\) is equal to ________. (rounded off to two decimal places)

Answer 1

Correct Answer: 0.49 - 0.51

To solve the given problem, we need to evaluate the double integral over the triangular region \(S\) with vertices at \((0, 0)\), \((0, \frac{\pi}{2})\), and \((\frac{\pi}{2}, 0)\). The double integral is given by \(\iint\limits_S\sin(x)\cos(y)dx\ dy\).

The triangular region can be described in the \(xy\)-plane with bounds \(0 \leq x \leq \frac{\pi}{2}\), and for each fixed \(x\), from \(y = 0\) to \(y = \frac{\pi}{2} - x\). Thus, the limits of integration are: \[ 0 \leq x \leq \frac{\pi}{2},\quad 0 \leq y \leq \frac{\pi}{2} - x. \]

We setup the integral as follows:

\[\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}-x}\sin(x)\cos(y)dy\ dx\]

First, integrate with respect to \(y\):

\[\int_0^{\frac{\pi}{2}-x}\cos(y)dy = \left[\sin(y)\right]_0^{\frac{\pi}{2}-x} = \sin\left(\frac{\pi}{2}-x\right)-\sin(0)\]

Since \(\sin\left(\frac{\pi}{2}-x\right)=\cos(x)\), this reduces to:

\(\cos(x) - 0 = \cos(x)\)

Now, substitute back into the integral with respect to \(x\):

\[\int_0^{\frac{\pi}{2}}\sin(x)\cos(x)dx\]

Using the identity \(\sin(2x) = 2\sin(x)\cos(x)\), rewrite the integral:

\[\frac{1}{2}\int_0^{\frac{\pi}{2}}\sin(2x)dx\]

Integrate with respect to \(x\):

\[\frac{1}{2}\left[-\frac{1}{2}\cos(2x)\right]_0^{\frac{\pi}{2}} = -\frac{1}{4}\left[\cos(\pi) - \cos(0)\right] = -\frac{1}{4}[-1 - 1]\]

This simplifies to:

\(-\frac{1}{4}(-2) = \frac{1}{2}\)

Thus, the value of the integral is \(\frac{1}{2} = 0.50\), which lies within the range \(0.49, 0.51\).

Therefore, the value is approximately 0.50.