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hi careless25
I have been looking at this and I realized that if we just square the actual integral and dont use the fomula given to us, we get to the correct answer. but this does:
Hi Bob bundy, now from here convert to polar coordinates EDIT: I am not sure what the integration limits for theta would be...if it is 0 to pi/2, then this works out. let u = r^2, du = 2rdr since we squared the original equation, we square root the final answer so This doesnt agree with what wolfram and wikipedia get . Any idea where I went wrong? Thanks
Hahaha I could not have thought of that on the final exam. Lost 10% right there .
Had me stuck too. But I found this:
Thanks bob bundy and bobbym. and told to use polar coordinates. I was confused on how to go from the above function to a function of x and y.
Hi Bob;
You could have chosen as the factorization. Now the result follows.
hi careless25 then Bob
Hi, 