Question

How do I evaluate this double integral, the integrand is $\sqrt{y}\sin y dydx$ ? ${{x}^{2}} < y < 1,0 < x < 1$

Answer 1

These problems are however a bit difficult to understand, but once understood and the concepts grasped, are pretty simple to solve. For sums like these we need to have a complete and clear understanding of double integrals and their representation in the three dimensional plane. We must also remember that the integrations in such problems can be reversed and the problem becomes simpler and yields the same answer. In this problem too, we are going to reverse the order of the integration, which means we will first integrate $dx$ over the region of $x$ and after that we are going to do it for $dy$ over the region of $y$.

Complete step by step solution:
Now we start off with the solution to the given problem by writing it as, we first reverse the integrals and say that,
$\int{\int\limits_{x=0}^{x=\sqrt{y}}{\sqrt{y}\sin ydydx}}$
$\Rightarrow \int{\sqrt{y}\sin ydy\int\limits_{x=0}^{x=\sqrt{y}}{dx}}$
Now we perform the integration over the inside one to get,
$\Rightarrow \int{\sqrt{y}\sin ydy\left( \sqrt{y}-0 \right)}$
Multiplying the result we get,
$\Rightarrow \int_{0}^{1}{y\sin ydy}$
Now, we apply integration by parts to the above equation to find the value of the respective integral. We do,

& \Rightarrow \left( -y\cos y \right)_{0}^{1}-\int\limits_{0}^{1}{\dfrac{d\left( y \right)}{dy}}.\left( -\cos y \right)dy \\\ & \Rightarrow \left( -\cos 1 \right)+\int\limits_{0}^{1}{\left( \cos y \right)dy} \\\ & \Rightarrow -\cos 1+\left( \sin y \right)_{0}^{1} \\\ & \Rightarrow -\cos 1+\sin 1 \\\ & \Rightarrow \sin 1-\cos 1 \\\ \end{aligned}$$ **Thus the answer to our double integral problem is $$\sin 1-\cos 1$$.** **Note:** For such types of problems, the most primary thing is to understand all the concepts related to double integration, its graphical representation, its application and other things. We must keep in mind of the little trick, that we can change the order of integration inside the double integral. This will yield the same answer, but will make the problem simpler. The final answer is obtained by doing a normal integral, first over ‘x’ and then over ‘y’. In general, the double integral represents an area in two dimensional space.