Find the volume of the solid generated by revolving the shaded region about the x-axis. 5x+2y =10
Hello and Welcome to OpenStudy! :) do you have a picture of where the shaded region would be?
Hi, and thanks!
I'm guessing you're in calculus?
Yes, Math 104
alright so if you rotate the shaded region around the x axis what do you think you would get? a washer? or cylinder thing?
Couldn't you just calculate the area of the triangle, and multiply that by\[2\pi r\]to get the volume? I hate shaded regions
Would that work Embryo? I hadn't thought of it... Hmm
And to jigglypuff314, I have this so far:
\[V \int\limits_{0}^{2} \pi (R(x))^2 dx\] \[R(x) = (10-5x)/2\] \[5x + 2(0) = 10\]
@Embryo I'm not getting the same answer as I would have if I did it the calculus way...
Then dangit, i really hate shaded regions
jghost, the first two steps look right to me I'm not sure what the third thing you have is :/ just plug in what you have for R(x) into the integral and solve
Sorry, the third thing was how I got the Upper Limit of Integration to be 2... but I'm having trouble solving :(
@Embryo, revolving the region would give you a cone. You could use the formula for a cone's volume to avoid the integration method.
Here's a non-calc way. By a little imagination, the volume of this thing is exactly the half of the volume of a cylinder with radius 5 and height 2.\[\pi \times 25 \times 2 \times \dfrac{1}{2}\]\[= 25\pi\]
Is that the right answer...?
\[V=\pi r^2\frac{ h }{ 3 }\]where r is the radius, being 5, and h= the height from the base to the tip, which is 2, like that @SithsAndGiggles ?
@Embryo yes :) that is the answer I got from the integration way
@Embryo, yeah
So @JGhost you now have multiple ways of solving this, pick your favorite and roll with it
\[\large \pi \int\limits_{0}^{2} (-2.5x + 5)^2~~ dx\] would be the integration way, which you can/should just plug into calculator :P
Hmm I went with the calc method and got 50pi/3 as the ans?
OK, then go with the calc method.
correct! :)
This makes way for another kind of curiosity. Why was the non-calc way wrong?
Ah, I see... never mind.
Sweet! The system accepted it as correct! Thanks peeps. @jigglypuff314 @Embryo @ParthKohli @SithsAndGiggles
glad we could help ^_^
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