How would you go about solving this integral (below)?

Question

2 (4 - x^2)^1/2
ò ò _ _ (x^2 + y^2 )^11/4 dy dx
0 0

I think you need to try changing to polar coordinates.


a. p(4/5)(2^1/2)
b. p(8/7)(2^1/2)
c. p(16/9)(2^1/2)
d. p(32/11)(2^1/2)
e. p(64/13)(2^1/2)
f. p(128/15)(2^1/2)

or is it none of these?

Answer 1

yes, use r = sqrt(x^2 + y^2)
then
(x^2+ Y^2) ^ 11/4 = r^ 11/2
dydx = rdrdt

Answer 2

Sorry, I cannot understand your notation.