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The region enclosed by : $y=6-x^2$ and $y=5$ I first get the inverse functions and the intersections and then work with the disk/washer method, the result is zero and I can't figure out what am I doing wrong, I can still solve it using the shell method. I can't upload a picture for the graph /my work(I get an error from the app)

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    $\begingroup$ Did you seriously ask us what you were "doing wrong" and then decline to show your work? Mathematicians are (often) really smart, but we're not psychic. $\endgroup$ – B. Goddard Apr 25 at 11:46
  • $\begingroup$ There is a guide here on how to write math on this site. If you write down your attempt at the washer method, we will actually be able to help. And while pictures (of graphs) are nice to have, they are not strictly necessary. $\endgroup$ – Arthur Apr 25 at 11:48
  • $\begingroup$ @B. Goddard where did I say this? It's clear that all that I needed is someone to do it again and it would be my job to figure out what wrong did I do. $\endgroup$ – user597368 Apr 25 at 12:33
  • $\begingroup$ @Arthur thank you for your suggestion. $\endgroup$ – user597368 Apr 25 at 12:35
  • $\begingroup$ No, you say you "can still do it." So the only thing that's clear is that you already can do the problem but somehow want us to guess what you did wrong. And this isn't a "do your homework" site. You don't just post problems and expect people to give you answers. $\endgroup$ – B. Goddard Apr 25 at 12:38
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If $y= 6- x^2$ then $x= \pm\sqrt{6- y}$. For any y between 5 and 6, we have a "washer" with inner radius $5- \sqrt{6- y}$ and outer radius $5+ \sqrt{6- y}$ so with area $\pi(5+ \sqrt{6- y})^2- \pi(5- \sqrt{6- y})^2= \pi(25+ 10\sqrt{6- y}+ 6- y)-\pi(25- 10\sqrt{6- y}+ 6- y)$$= 20\pi\sqrt{6- y}$.

Integrate that, with respect to y, from y= 5 to y= 6.

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  • $\begingroup$ Thank you for helping. Compared this to what I did, and figured out where was I wrong. $\endgroup$ – user597368 Apr 25 at 12:31

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