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anonymous
 one year ago
The region between the graph f(x)=x^(1/2) and the xaxis for x∈[0,16] is revolved about the line y=8. Which integral will determine the volume of the solid that is generated?
I got π(x64)dx, x∈[0,16] using the yA=1/2[(f(x))^2  (g(x))^2]dx x∈[a,b] equation, but the answer is ∫π[64−(8−x^(1/2))^2]dx x∈[0,16]. No idea where I went wrong, but maybe this isn't the right equation to use?
anonymous
 one year ago
The region between the graph f(x)=x^(1/2) and the xaxis for x∈[0,16] is revolved about the line y=8. Which integral will determine the volume of the solid that is generated? I got π(x64)dx, x∈[0,16] using the yA=1/2[(f(x))^2  (g(x))^2]dx x∈[a,b] equation, but the answer is ∫π[64−(8−x^(1/2))^2]dx x∈[0,16]. No idea where I went wrong, but maybe this isn't the right equation to use?

This Question is Closed

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1are we using shells or discs?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1shells add the area of rectangles: 2pi r discs add the area of circles: pi r^2

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1it looks like disc method to me

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442770213833:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is a problem semirelated to centroids. I assumed it was Disc Method, but either I did something wrong or I'm overthinking everything.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1we have a circle of radius 8, that have a center removed from them of a radius of 8sqrt(x)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1pi (8^2  (8sqrt(x))^2) is the area of a disc for a given value of x

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1it looks like you have an error in your application. can you write it out in more detail? what were your thoughts?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1washer method is another name for removing a disc out of a disc ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was looking at this in terms of centroids. Most of the equations are the same as the ones for volume. I actually have the Washer Method equation down (I just realized PFFT). The Disc Method equation is just V=πr^2. The way your wrote it as π[8^2  (8x^(1/2))^2] is because of the rotation around the y=8, right?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1yes we have a cylindar, or radius 8 (center at y=8, and edge at y=0) we are gutting out the middle of it, so at an given x value, we are 8, minus sqrt(x) dw:1442770823342:dw since sqrt(x) + r = 8

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1pi(R^2  r^2) gives us the portion that we are concerned with.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you very much for you answers~!
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