y= -2x+ 4= 0 (crossing the x-axis) when x= 2 so x ranges from 1 to 2.
It is NOT a cone, it is "frustrum" of a cone, not including the "point".
I tried to post this before, but I don't think it went through!
2. Alright, so I first drew it out, and the shape is a cone.
Since it revolves around a y-axis, the limits should be also in terms of y.
You mean around a line parallel to the x-axis.
Every cross-section is a circle.
Yes, having radius x- 1. Since y= -2x+ 4, 2x= 4- y, x= 2- y/2 and x- 1= 2- y/2- 1= 1- y/2. The area of such a circle is \pi(1- y/2)^2and taking its thickness to be "dy", its volume is \pi(1- y/2)^2dy= \pi (1- y+ y^2/4)dy.<br />
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So this is how I wrote it out:<br />
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∫0 to 4 of ∏[((-y/2) + 1)^2]dy<br />
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∫0 to 4 of ∏[(y^2/4) -y + 1]
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</blockquote> You forgot the "dy" but no matter.<br />
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= 1/4 * (y^3)/3) - (y^2)/2) + x |0 to 4<br />
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= ((y^3)/12) - 1/2(y^2) + x |0 to 4
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</blockquote>and you really mean "y", not "x" there. But again, you put y= 4 into that so, no matter!<br />
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= ((4)^3)/3 - 1/2(4)^2 + 4 -(0)
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</blockquote>And here, you should have "(4)^3/12" not "(4)^3/3"<br />
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= 64/12 - 8 + 4
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</blockquote>64/12= 16/3 so this is 16/3- 24/3+ 12/3= (26- 24)/3= 4/3.<br />
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= 1.33 or 4/3<br />
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So... Is that right? <br />
Thank you so much for checking my work! :)
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</blockquote> Yes, that looks good to me. The fact that it is only a part of a cone is not relevant.