Question #91c37

1 Answer
Feb 11, 2018

The outer integral results in an indeterminate form.

Explanation:

The indefinite integral

#int y^2sqrt(x^2+y^2)dy#

is done by letting #y = xtan(u)# and #dy = xsec^2(x)du#

I will not do the integration but I will give you the results provided by WolframAlpha :

#int y^2sqrt(x^2+y^2)dy = 1/8 (y sqrt(x^2 + y^2) (x^2 + 2 y^2) - x^4 ln(sqrt(x^2 + y^2) + y)) + c#

We evaluate the the right side of the above at #y = sqrt(a^2-x^2)#

#1/8 (sqrt(a^2) sqrt(a^2 - x^2) (2 a^2 - x^2) - x^4 ln(sqrt(a^2 - x^2) + sqrt(a^2)))#

Subtract the right side of the above evaluated at #y = 0#:

#1/8 (sqrt(a^2) sqrt(a^2 - x^2) (2 a^2 - x^2) - x^4 ln(sqrt(a^2 - x^2) + sqrt(a^2))+x^4ln(x))#

I used WolframAlpha to do the indefinate integration of the above with respect to x:

#1/8 int (sqrt(a^2) sqrt(a^2 - x^2) (2 a^2 - x^2) - x^4 ln(sqrt(a^2 - x^2) + sqrt(a^2))+x^4ln(x)) dx = 1/40 (6 (a^2)^(3/2) x sqrt(a^2 - x^2) - x^5 ln(sqrt(a^2 - x^2) + sqrt(a^2)) - sqrt(a^2) x^3 sqrt(a^2 - x^2) + 4 a^4 sqrt(a^2) tan^(-1)(x/sqrt(a^2 - x^2)) + x^5 ln(x)) + c#

Please observe that, if we attempt to evaluate the right side of the above at #x =a#, it will cause a division by 0 within the inverse tangent function. Therefore the definite integral is indeterminate