Question

Evaluate the iterated integral. (Note that it is necessary to switch the order of integration.) $$ \int_{0}^{2} \int_{\frac{1}{2} x^{2}}^{2} \sqrt{y} \cos y d y d x $$

Short answer

The evaluated integral is \(2\sin 2 + \cos 2 - 1\).

Step-by-step solution

Decide on the order of integration

The limits of \(y\) are functions of \(x\) and the limits of \(x\) are constants. The integral is simpler if we make \(y\) the outer integral and \(x\) the inner integral. This is because the function we are integrating, \(\sqrt{y} \cos y\), is a function of \(y\) only.

Change the order of integration

The region can be described as: \[0 \leq x \leq 2, \frac{1}{2}{x^{2}} \leq y \leq 2\]To change the order of integration, we solve for \(x\) from the inequalities \(\frac{1}{2}{x^{2}} \leq y \leq 2\) and obtain the region as \(0 \leq y \leq 2\) and \(0 \leq x \leq \sqrt{2y}\). The integral then changes to \[\int_{0}^{2} \int_{0}^{\sqrt{2y}} \sqrt{y} \cos y d x d y\].

Calculate the inner integral

Solving this inner integral first which is with respect to \(x\), and because the function inside is not a function of \(x\), the integral simplifies to \(x\sqrt{y}\cos y\). When we apply the limits \(0\) and \(\sqrt{2y}\) we get: \[\sqrt{2y^{2}}.\cos y - 0\]

Calculate the outer integral

Now we solve the outer integral:\[\int_{0}^{2} y \cos y dy\]This can be solved using the method of integration by parts: if we have an integral in the form \(\int u dv\), it can be solved by \(uv - \int v du\). Let \(u = y\) and \(v = \sin y\), then the integral simplifies to: \[2\sin 2 - \int_{0}^{2} \sin y dy = 2\sin 2 - [-\cos y]_{0}^{2} = 2\sin 2 +\cos 2 -1.\]