So, I want to use Washer method to find $$\pi \int_{y=0}^2 (\sqrt{4y})^2-(\sqrt{4y}-0.1)^2 dy.$$

I've literally spent a day trying to do so, and when I treat each part separately and then subtract the answer keeps resulting in a) a negative or b) two drastically far apart numbers

Hence, I think something is wrong in the initial set up. I'm trying to model rotating this line: about the y axis. Any thoughts on what is wrong?

the line im trying to model- desmos graph

  • $\begingroup$ I just tried it again. Seems fairly straightford to me at first, but then my result was 0.06205cm-- which seems impossible given that the distance between the lines is 0.1cm for the majority of its length $\endgroup$ – Rae Ann Aug 24 at 10:05
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    $\begingroup$ The washer method is a mnemonic for setting up a correct integral when finding the volume of a solid of revolution. Once you have an integral (which you seem to have here), the washer method has done its job, and all that's left is pure calculation. $\endgroup$ – Arthur Aug 24 at 10:12
  • $\begingroup$ Okay. must be some problem in the calculation. $\endgroup$ – Rae Ann Aug 24 at 10:13
  • $\begingroup$ It should be $\pi(0.4*2^{3/2}*2/3-0.02)$. Pay attention to the position of the factor $0.4$. $\endgroup$ – Robert Z Aug 25 at 8:25

From your graph it seems that you are considering the solid obtained by rotating the curve about the $y$-axis. Is it correct?

Why don't you simplify before integrating? $$(\sqrt{4y})^2-(\sqrt{4y}-0.1)^2=0.4 \sqrt{y}-0.01.$$ It remains to evaluate $$\pi\left(0.4 \int_0^2 \sqrt{y}\,dy -0.01 \int_0^2 \,dy\right)= \pi\left(0.4 \left[\frac{2y^{3/2}}{3}\right]_0^2 -0.01 \cdot 2\right).$$ What is the final result?

  • $\begingroup$ Sorry where did the last ∫1dy come from? $\endgroup$ – Rae Ann Aug 24 at 10:18
  • $\begingroup$ I think the result is 2.3595..., does that seem correct to you? Thank you! $\endgroup$ – Rae Ann Aug 24 at 10:20
  • $\begingroup$ I got $2.3067$. You may write simply $∫dy$. BTW may I see the original statement of your problem? $\endgroup$ – Robert Z Aug 24 at 10:35
  • $\begingroup$ Sorry I don't really have an original statement of the problem-- I was trying to model the volume of material needed for a parabolic solar cooker (though this one is off as it's narrow at the base), but the values chosen are a bit arbitrary. The focal point was 1, so equation $y=x^2/4$ came from there, then decided to do it around y axis so rearranged to get current equation. The -0.1 is just because the average solar cooker is 1mm thick, so the second function needed to be that distance away. So pretty much the integral set up in the question, and the graph, is all I have sorry. $\endgroup$ – Rae Ann Aug 24 at 10:44
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    $\begingroup$ We have that $\int y^{1/2} dy= \frac{y^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c$ and $\int_a^b dy=b-a$. $\endgroup$ – Robert Z Aug 24 at 15:33

If you're rotating about the $x$-axis, you need functions in terms of $x$ for the first quadrant:

$$ x=\sqrt{4y}\implies y=\frac{x^2}{4},\\ x=\sqrt{4y}-0.1\implies y=\frac{(x+0.1)^2}{4}. $$

Your integral should look like this where $2$ is supposedly the upper bound of integration:

$$ V=\pi\int_{0}^{2}\left[\left(\frac{x^2}{4}\right)^2-\left(\frac{[x+0.1]^2}{4}\right)^2\right]\,dx. $$

If you're rotating about the $y$-axis, then your integral looks fine. There must have been a mistake in your calculations:

$$ V=\pi\int_{0}^{2}\left[\left(\sqrt{4}y\right)^2-\left(\sqrt{4y}-0.1\right)^2\right]\,dy=\\ \pi\int_{0}^{2}\left[4y-\left(4y-0.2\sqrt{4y}+0.01\right)\right]\,dy=\\ \pi\int_{0}^{2}\left(4y-4y+0.2\cdot 2\sqrt{y}-0.01\right)\,dy=\\ \pi\int_{0}^{2}\left(0.4\sqrt{y}-0.01\right)\,dy=\\ 0.4\pi\int_{0}^{2}\sqrt{y}\,dy-0.01\pi\int_{0}^{2}\,dy=\\ 0.4\pi\left(\frac{2\sqrt{2^3}}{3}-\frac{2\sqrt{0^3}}{3}\right)-0.01\pi(2-0)=\\ 0.8\pi\frac{\sqrt{2^2\cdot 2}}{3}-0.02\pi=\\ 0.8\pi\frac{2\sqrt{2}}{3}-\pi\frac{0.02\cdot 3}{3}=\\ \pi\left(\frac{0.8\cdot 2\sqrt{2}}{3}-\frac{0.06}{3}\right)=\\ \pi\frac{1.6\sqrt{2}-0.06}{3}\approx 2.31\ \text{cubic units}. $$

This is easy to integrate. Just remember that $\int\sqrt{x}\,dx=\frac{2\sqrt{x^3}}{3}+C$ and $\int\,dx=x+C$.

  • $\begingroup$ Apologies, i think i mistyped x rather than y. $\endgroup$ – Rae Ann Aug 24 at 10:14
  • $\begingroup$ Sorry I'm a bit confused about what's going on in the 3rd to 4th lines-- if you change $ \sqrt (4y)$ to $2(\sqrt y)$, where does the $\sqrt 4y$ come from in line 4? $\endgroup$ – Rae Ann Aug 24 at 13:15
  • $\begingroup$ Sorry. I forgot to remove the $4$ when I was copying things. I fixed it now. $\endgroup$ – Michael Rybkin Aug 24 at 13:35
  • $\begingroup$ Also why is $\pi$ both a constant and after 0.01 in the 5th line? and where does the 0.06 come from? $\endgroup$ – Rae Ann Aug 24 at 14:12
  • $\begingroup$ I edited the answer. Take a look at it again. $\endgroup$ – Michael Rybkin Aug 24 at 14:38

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