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[tex] \int_{0}^{\sqrt{\pi}} ( \int_{x}^{\sqrt{\pi}} sin y^2 dy) dx [/tex]

now i know i have to change the order of this

the integrand is bounded by the triangle from x = 0 to [itex] x= \sqrt{\pi} [/itex] 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!