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I have been working on this sum for a while. The question asks to evaluate the double integral. $$\int_0^1\int_p^1 \frac {x^3}{\sqrt{1-y^6}} dydx$$ where $p$ is equal to $x^2$. I know that I have to solve the $y$ integral first and then the $x$. But I don't know how to solve the root integral. Applying the formula $$\int \frac{1}{\sqrt{1-t^2}}dt$$ where $t=y^3$ isn't working. Any ideas as to how I must proceed with the integral? Once I get the integral, I must substitute the limits and then the integral would be in terms of $x$ and I must integrate it. Am I correct?

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Change the order of integration. You can do this by drawing the region of integration and seeing that the integral is just

$$\int_0^1 \frac{dy}{\sqrt{1-y^6}} \, \int_0^{\sqrt{y}} dx \, x^3$$

which is

$$\frac14 \int_0^1 dy \frac{y^2}{\sqrt{1-y^6}}$$

or

$$\frac{1}{12} \int_0^1 \frac{du}{\sqrt{1-u^2}} = \frac{\pi}{24}$$

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  • $\begingroup$ Ah! Ok I need to draw the sketch. But how do I know if my sketch makes any sense at all? I am not very good at graphing. Any tips? $\endgroup$ – Artemisia Nov 10 '13 at 12:29
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    $\begingroup$ @Artemisia: you can draw $y=x^2$, right? And you can look sideways...right? At least these were explicit prerequisites for Calc III at my university. That and familiarity with the number $7$. $\endgroup$ – Ron Gordon Nov 10 '13 at 12:31
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    $\begingroup$ @Artemisia: usually for problems like this, the sketches are no more complicated than this. But in general, just make sure that the region is convex and makes sense sideways. $\endgroup$ – Ron Gordon Nov 10 '13 at 12:50
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    $\begingroup$ @GerryMyerson: amazon.com/Penguin-Book-Curious-Interesting-Numbers/dp/… $\endgroup$ – Ron Gordon Sep 10 '15 at 23:58
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    $\begingroup$ Thanks, Ron. I also found this: imdb.com/title/tt0054047 $\endgroup$ – Gerry Myerson Sep 10 '15 at 23:59

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