# Iterated Integrals

i need to solve this
$$\int_{0}^{\sqrt{\pi}} ( \int_{x}^{\sqrt{\pi}} sin y^2 dy) dx$$

now i know i have to change the order of this
the integrand is bounded by the triangle from x = 0 to $x= \sqrt{\pi}$ here's where i am stuck
what is the boundary of the y?? is y bounded below by x=0 and above by x =1??

so what would the limits of integration change to?? (for the inside one from 0 to root pi?) and the outside one stays the same??

pelase help!

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ehild
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stunner5000pt said:
i need to solve this
$$\int_{0}^{\sqrt{\pi}} ( \int_{x}^{\sqrt{\pi}} sin y^2 dy) dx$$

now i know i have to change the order of this ....

Look at the attached figure. You have to integrate for the yellow triangle, according to the boundaries of your original integral. If you change the order of integration, that is you integrate by x first, it would go from x=0 to x=y; and then by y which goes from y=0 to y= sqrt (pi).

ehild

Last edited:
ehild said:
Look at the attached figure. You have to integrate for the yellow triangle, according to the boundaries of your original integral. If you change the order of integration, that is you integrate by x first, it would go from x=0 to x=y; and then by y which goes from y=0 to y= sqrt (pi).

ehild
how do you know that it is the upper triangle and not the lower triangle??

Because in the original integral:

$$\int_{0}^{\sqrt{\pi}} ( \int_{x}^{\sqrt{\pi}} sin y^2 dy) dx$$

Look at the limits for the dy integral, y goes from y=x to y=root pi. If it were the white triangle in the image, then y would be going from 0 to x.

ehild is completely correct

marlon

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