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I have to solve the following definite integral $$\int_{0}^{4}r^3 \sqrt{25-r^2}dr=3604/15$$

I have tried a change of variables given by $u=\sqrt{25-r^2}$ where then I find that $dr=-u du/r$ and $r^2=25-u^2$. That change of coordinates then give the following integral $$-\int_{3}^{5}(25-u^2)u du$$

However, it now evaluates to $-3604/15$ and I wonder why I have a negative popping up if I have done what seems like the correct thing to do.

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    I made the Same mistake earlier today You changed the bounds that gives another minus –  May 15 '16 at 21:57
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    the limits should be from 5 to 3 - lower limit is $\sqrt{25-0}=5$ – WW1 May 15 '16 at 21:58

4 Answers4

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You have the lower and upper bounds of integration mixed up, they should be switched.

Then the property $\int^a_b f(x)dx=-\int^b_a f(x)dx$ should fix the issue for you.

ASKASK
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  • It is allowed to put a smaller bound on the top? I thought I had to put them in order. Thanks –  May 15 '16 at 21:59
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    It is perfectly allowed, in fact you must make the bounds according to what the suggestion requires. In particular r=0 corresponds to u=5 so the lower bound must be 5 – ASKASK May 15 '16 at 22:01
  • Thank you for your answer! –  May 15 '16 at 22:02
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If you do $u=\sqrt{25-r^2}$, you have $$ du=-\frac{r}{\sqrt{25-r^2}}\,dr $$ so $u\,du=-r\,dr$. So your computation is right, up to this point. Next, $r^2=25-u^2$, and still good.

However, $r=0$ gives $u=5$ and $r=4$ gives $u=3$, so the integral is $$ \int_5^3 -u(25-u^2)\,du $$

egreg
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The lower limit should be $5$ and the upper should be $3$. Switching them introduces a minus sign, making the result positive.

Ross Millikan
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You've already been answered about the confusion with the limits. Now you can try the following and not make a substitution and thus not change the limits. Integrate by parts:

$$\begin{cases}u=r^2&u'=2r\\{}\\v'=r\sqrt{25-r^2}&v=-\frac13(25-r^2)^{3/2}\end{cases}\;\;\implies$$$${}$$

$$\int_0^4r^3\sqrt{25-r^2}\,dr=\left.-\frac{r^2}3(25-r^2)^{3/2}\right|_0^4+\frac13\int_0^2(2r\,dr)(25-r^2)^{3/2}=$$

$$=-\frac{16}3\cdot27-\left.\frac13\frac25(25-r^2)^{5/2}\right|_0^4=-\frac{27\cdot16}3-\frac2{15}\left(243-3125\right)=$$

$$=-\frac{432}3+\frac{5764}{15}=\frac{3604}{15}$$

DonAntonio
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