# Thread: Evaluate

1. ## Evaluate

semicircle bounded by x-axis and the curve y=sqrt(1-x^2)
double int[e^((x^2)+(y^2))]dydx

2. Originally Posted by chevy900ss
semicircle bounded by x-axis and the curve y=sqrt(1-x^2)
double int[e^((x^2)+(y^2))]dydx

Polar coordinates: $x=r\cos\theta\,,\,y=r\sin\theta\,,\,\,0\leq \theta\leq \pi\,,\,\,0\leq r\leq 1$, so:

$\int\limits_0^{\pi}d\theta \int\limits_0^1re^{r^2}dr=$...

Tonio

3. is the next step?
int(0topi)1/3edtheta

4. Originally Posted by chevy900ss
is the next step?
int(0topi)1/3edtheta

$\int re^{r^2}dr=\frac{1}{2}e^{r^2}$.
I can't understand what you wrote.

Tonio

5. yeah tried to skip steps but, where did the r^2/2 go? I thought it would have been = to ((1/2)r^2)e^r^2

6. Originally Posted by chevy900ss
yeah tried to skip steps but, where did the r^2/2 go? I thought it would have been = to ((1/2)r^2)e^r^2

What kind of question is that? Derivate my solution and check whether you get what's inside of the integral or not!

tonio

7. ok ill start over, i didnt mean to sound like i was right and you were wrong. sorry if i came off that way. I really barely have a clue what im doing. i got that the integral of [re^(r^2)]dr equals [(r^2/2)e^(r^2)].

8. Originally Posted by chevy900ss
ok ill start over, i didnt mean to sound like i was right and you were wrong. sorry if i came off that way. I really barely have a clue what im doing. i got that the integral of [re^(r^2)]dr equals [(r^2/2)e^(r^2)].

Well that's false, and checking it is very simple: derivate the "solution" and you'll find out that it doesn't equal the integrand!
One MUST know derivation very-very well to go into antiderivation/indefinite integration, otherwise things get tough.

Tonio