How does this integral come out to be zero?

[hivesaeed4]

The question is:
∫ (y(4-x^2-y^2)^(3/2) dy with y=(-sqrt(4-x^2)) to sqrt(4-x^2)) dy

I evaluated it online and it came out to be zero which was quite astonishing since both limits of y are not the same. The limits are the additive inverses of each other. Does this mean that regardless of the integrand if the limits are additive inverses of each other the integral will evaluate to be zero?

 

[DonAntonio]

The question is:
∫ (y(4-x^2-y^2)^(3/2) dy with y=(-sqrt(4-x^2)) to sqrt(4-x^2)) dy

I evaluated it online and it came out to be zero which was quite astonishing since both limits of y are not the same. The limits are the additive inverses of each other. Does this mean that regardless of the integrand if the limits are additive inverses of each other the integral will evaluate to be zero?

I really don't understand what you mean by "$y=-\sqrt{4-x^2}\,\, to\,\, \sqrt{4-x^2}\, dy$" , but

$\int y\left(4-x^2-y^2\right)^{3/2}\,dy=-\frac{1}{2}\int -2y\left(4-x^2-y^2\right)^{3/2}\, dy=\frac{2}{5}\left(4-x^2-y^2\right)^{5/2}+C\,,\,\,C=$ constant in y .

DonAntonio

 

[hivesaeed4]

My bad. The limits of y were -sqrt(4-x^2) to sqrt(4-x^2)).

(And by the way you forgot to multiply the -1/2 factor in your last step).

 

[DonAntonio]

My bad. The limits of y were -sqrt(4-x^2) to sqrt(4-x^2)).

(And by the way you forgot to multiply the -1/2 factor in your last step).

Indeed I did, but then we're done as your integral equals

$-\frac{1}{5}\left[\left(4-x^2-(\sqrt{4-x^2})^2\right)-\left(4-x^2-(-\sqrt{4-x^2})^2\right)\right]=0$ , and there's anything surprising in this as the integrand

function is an even one (in y, of course) and thus it vanishes in any symmetric interval.

DonAntonio

 

[hivesaeed4]

Oh. Oh I see. Thanks DonAntonio.