$\int_{x=0}^{x=2}$$\int_{y=x^2}^{y=4}$$(\frac{ x^3}{\sqrt {x^4+y^2}})$$dydx$

This is the integral I have been given to solve. I've already sketched the domain and reversed the limits which gives:

$\int_{y=0}^{y=4}$$\int_{x=0}^{x=\sqrt y}$$(\frac{ x^3}{\sqrt {x^4+y^2}})$$dxdy$

When it comes to the substitution and finding the new limits with variable u, I'm getting a bit confused. I'm assuming that $u=\sqrt {x^4+y^2}$ but I'm not sure what to do from there.


I thought that it might be instructive to present a solution using a slightly different substitution.

The new inner integral $I$ becomes

$$I=\int_0^\sqrt {y} \frac{x^3}{\sqrt{x^4+y^2}}\,dx$$

Now, if we simply enforce the substitution $x^4=u$, then $4x^3\,dx=du$ and the limits of integration begin at $=-0$ and end at $u=y^2$. Therefore, we can write

$$I=\frac14\int_0^{y^2} \frac{1}{\sqrt{u+y^2}}\,du=\frac12 \left(\sqrt{2y^2}-\sqrt{y^2}\right)$$

Can you finish now?

  • $\begingroup$ Yes, I can. Thank you so much for your help. $\endgroup$ – Sophie Meaden Jan 18 '16 at 23:09
  • $\begingroup$ You're welcome Sophie! My pleasure and pleased that I could help. - Mark $\endgroup$ – Mark Viola Jan 18 '16 at 23:20
  • $\begingroup$ Cheers again, this exact question came up in my Finals exam this morning. 15 marks in the bag because of you, you kind soul! $\endgroup$ – Sophie Meaden Jan 19 '16 at 16:09
  • $\begingroup$ Sophie, that's great! Congratulations! $\endgroup$ – Mark Viola Jan 19 '16 at 16:15

As you have, $u=\sqrt{x^4+y^2}\;\;$ so $du=\frac{4x^3}{2\sqrt{x^4+y^2}} dx$ and you have



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