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

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 $9$.

Arthur
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Umair Tahir
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  • What are your intermediate calculations? Maybe if you supply those, we can tell you what your mistake was. – Arthur Jan 13 '20 at 12:02
  • evaluated integral without limits is $-(((9-y^2)^(3/2))/3)$

    when i put $0>y>3$ then the answer will be $0$ instead of $9$

     – Umair Tahir Jan 13 '20 at 12:16

1 Answers1

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

The Pointer
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  • Thanks for sharing this answer. But can you please tell me how you have change $0>x>3$ to $9>x>0$ – Umair Tahir Jan 13 '20 at 12:11
  • @UmairTahir See that we implemented the change of variable $u = 9 - y^2$? When we perform a change of variable, the bounds of integration must also change to account for this change of variable. So to find the new bounds of integration, we substitute the old bounds $u = 3$ and $u = 0$ into the formula for the change of variable, $u = 9 - y^2$, to get $u = 9 - 3^2 = 0$ for the first bound, and $u = 9 - 0^2 = 9$ for the second bound. – The Pointer Jan 13 '20 at 12:14
  • @UmairTahir See here: https://en.wikipedia.org/wiki/Change_of_variables#Integration It is often referred to as "$u$-substitution": https://en.wikipedia.org/wiki/Integration_by_substitution – The Pointer Jan 13 '20 at 12:15
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    Thanks a lot. You don't know how grateful of you I am. – Umair Tahir Jan 13 '20 at 12:19