Integrate I= $2\pi\int_0^2 y(4-y^2) dy$

by Bhumi Patel   Last Updated March 14, 2019 20:20 PM

Consider the integral $$I= 2\pi\int_0^2 \! y(4-y^2) \, \mathrm{d}y$$

When I try integrating this, first by distributing the $y$, I get my answer as 8$\pi$ and then using Chain Rule in Reverse my answer is $0$. How is that possible?

Edit : Using Chain Rule in Reverse: When I integrate it using chain rule in reverse, which is an easier way to think of this by manipulating the integral in a way that the differential of the inside is outside. So it would be: = $2\pi\int_0^2 y(4-y^2) dy$ = -pi/2 $(4-y^2)$ and when you put in the limit of 2 and 0 you get the answer of the integral as 0. You can cross check by differentiating -pi/2 $(4-y^2)$ and see it equals to $2\pi y(4-y^2)$

Answers 3

$$ 2\pi\int_0^2 y(4-y^2)\,dy= 2\pi\int_0^2 (4y-y^3)\,dy= 2\pi\left[4\frac{y^2}{2}-\frac{y^4}{4}\right]_{0}^{2}=\\ 2\pi\left(2\cdot 2^2-\frac{2^4}{4}-(0-0)\right)=2\pi(8-4)=8\pi $$

That is the correct answer.

I don't really understand what you mean by "using the chain rule in reverse". But if you differentiate your answer, you will, of course, get zero because the derivative of a constant ($8\pi$ is a constant) is zero:

$$ C'=0 $$

The definite integral is always a number. It's not the same thing as the indefinite integral (the antiderivative of a function). If you integrate a function (that's when there are no limits of integration) and then differentiate the result that you get, you will get back your original function. But, as I said, that's called indefinite integration.

Edit: It looks like by the chain rule in reverse, you mean u-substitution. Let's try it: $$ 2\pi\int_0^2 y(4-y^2)\,dy= -\frac{2}{2}\pi\int_0^2(4-y^2)\frac{d}{dy}(4-y^2)\,dy=\\ -\pi\int_4^0u\,du=-\pi\frac{u^2}{2}\bigg|_4^0=-\pi\left(\frac{0^2}{2}-\frac{4^2}{2}\right)=-\pi(0-8)=8\pi $$ The answer is the same. You probably forgot to change the limits of integration.

Michael Rybkin
Michael Rybkin
March 14, 2019 19:51 PM

I am unsure what you mean by "using the Chain Rule in Reverse." Since you did not provide any of your work, I do not know how to help you with your approach to this problem.

Instead, I will present an alternative method for evaluating your integral with the hope that it will prove helpful. Start withthe substitution $u= y^2$. Your integral then becomes quite simple to evaluate: $$2\pi\int_0^2 \! y(4-y^2) \, \mathrm{d}y = \pi\int_0^4 \! (4-u) \, \mathrm{d}u = \pi\left[4u -\frac{1}{2}u^2\right]_0^4 = \pi \left(16 - \frac{1}{2} \cdot16\right) = 8\pi.$$

This solution is confirmed by Wolfram Alpha.

March 14, 2019 19:57 PM

Let $t= 4-y^2$ then $dt= -2ydy$ so

$$I=-\pi\int_4^0 \! t \, \mathrm{d}t =\pi\int_0^4 \! t \, \mathrm{d}t =\pi{t^2\over 2}\mid^4_0 = 8\pi$$

Maria Mazur
Maria Mazur
March 14, 2019 20:15 PM

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