I am unable to find correct answer after putting limits

40 Views Asked by At

$$ \int_0^3\int_0^{\sqrt{9-y^2}}y\,dx\,dy $$

I am able to evaluate the integral correct but when i put the limits on the last integral my calculation returns $0$, but the correct answer should be $9$.

1

There are 1 best solutions below

4
On BEST ANSWER

$$\begin{align} \int_0^3 \int_0^{\sqrt{9 - y^2}} y \ dx dy &= \int_0^3 [yx]^{\sqrt{9 - y^2}}_0 \ dy \\ &= \int_0^3 y\sqrt{9 - y^2} \ dy \\ &= \dfrac{-1}{2} \int_9^0 \sqrt{u} \ du \ \ \text{(Let $u = 9 - y^2 \Rightarrow \dfrac{du}{-2y} = dy$.)} \\ &= \dfrac{-1}{2} \left[ \dfrac{2u^{\frac{3}{2}}}{3} \right]^0_9 \\ &= \dfrac{-1}{2} \left[ 0 - \dfrac{2(27)}{3} \right] \\ &= \dfrac{27}{3} \\ &= 9 \end{align}$$