Question

Evaluate the double integral:

\[
\int_{0}^{9}\int_{\sqrt{y}}^{3}sin(\pi x^{3})\; dx\, dy
\]

Answer 1

The first step is usually to evaluate the "inside integral" first:

\[\huge \int_{x=\sqrt{y}}^{x=2}\sin(\pi x^3)\] dx

But since that doesn't have any friendly answer.. you may want to try changing order of integration.

Answer 2

You cant make that integral, it leaves you with imaginary parts, and gamma function, and if not wrong with eliptical.

Answer 3

First, perform the inner most integrand. \[\int\limits_{\sqrt{y}}^{3}\sin(\pi x^3) dx\]Let's solve by u-substitution. \[\text{Let}~~ u = x^3 \rightarrow {du \over dx} = 3x^2\]We need to change our limits of integration to be \[x:y^{3/2} \rightarrow 27\]Therefore, the integral becomes \[{1 \over 3}\int\limits_{y^{3/2}}^{27} \sin(\pi u) du\]Evaluating this becomes \[-{1 \over 3}(\cos(27\pi) - \cos(\pi y^{3/2})) \]The outer integrand becomes \[-{1 \over 3}(\cos(27 \pi)y + \int\limits_0^9\cos(\pi y^{3/2}) dy \]Again, by u-substitution \[\text{Let} ~~ u = y^{3/2} \rightarrow {du \over dy} = {3 \over 2} y^{1/2}\]Again, let's change our limits of integration to \[y:0 \rightarrow 27\]Substituting in u, the integral becomes \[-{1 \over 3} \left( \cos(27 \pi) y |_0^{9} + \int\limits_0^{27} \cos(\pi u) du \right)\]Solving, we get \[-{1 \over 3} \left( \cos(27\pi)y |_0^{9} + \sin(\pi y) |_0^{27} \right)\]

Answer 4

Thank you but I don't think the u substitution is correct. You end up with \[du=3x^2dx\] but notice that 3x^2 is together with dx, and that form is not in the original integrand. It seems like you used \[\frac{du}{3}=x^2dx\] which would make the integrand have a term of x in it, therefore rendering the u substitution not possible.

Please correct me if I'm wrong. This has been a challenging one for me to solve.

Answer 5

Change the order of integration. Then you will have something which is much more manageable:

Here are some very good notes on how to do this:
http://mathinsight.org/double_integral_change_order_integration_examples
http://tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx
http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/part-a-double-integrals/session-49-exchanging-the-order-of-integration/ (click example to see the PDF doc)

Answer 6

Thanks @mathteacher1729, changing the variables made it easy as pi :D. I was having trouble visualizing the region, it was simply y=x^2. The links you gave are very helpful. Thanks a lot.