<p>^^ycm1957</p>
<p>For a) could it also be: ∫(0→9) 6 - 2(x)^(1/2) dx</p>
<p>for those of you who are afraid you didn’t get a five because of some silly errors - don’t lose hope! you can still pick up several points just for setting things up correctly and having the right units. and you don’t even need all the points to get a five - really you only need like half on each question.</p>
<p>This is shocking. I got the (c) and (d) parts right for each frickin’ question, but bombed the (a) and (b).</p>
<p>At least they give points for setup.</p>
<p>
His b is wrong because he forgot to include pi. His a is right, he just took a different integral than you did.</p>
<p>
how?
is it not top - bottom?</p>
<p>^ Notice he took the integral from 0 to 6 of (y^2)/4 dy. </p>
<p>That will produce the same value as the integral from 0 to 9 of (6 - 2x^(1/2)) dx</p>
<p>i believe 4B should be:</p>
<p>b) pi∫(0→9) (7-2x^(1/2))^2</p>
<p>For 5. C) did you guys get x = 3 and it not being a max. or a min. I just remember I guessed on that one…</p>
<p>
That would only produce a disk. You need a washer to account for the space between y=7 and y=6.</p>
<p>The correct answer is pi∫(0→9) (7-2x^(1/2))^2 - (7-6)^2 dx</p>
<p>
We havn’t covered 5 yet but yes x = 3 is neither a relative maximum nor minimum. However x = 2^(1/) is a relative maximum.</p>
<p>I think i got 4b right. I just forgot to type pi here.</p>
<p>No, the correct answer is pi∫(0→9) ((7-6)^2) - (7-2x^(1/2))^2)dx </p>
<p>Top - Bottom.</p>
<p>^^^Jersey, how did you get that as a relative max.? I completely missed that… </p>
<ol>
<li>(b) x = -2, 2, 3</li>
</ol>
<p>
That would be the case if you were revolving around the x-axis, but you are revolving around y= 7.</p>
<p>let’s move on to 5 now.</p>
<p>You’re still revolving about a horizontal line, though.</p>
<p>h(x) = g(x) - 1/2x^2</p>
<p>h’(x) = g’(x) - x</p>
<p>so g’(x) and x should be the same, meaning that (2^(1/2),2^(1/2)) and (3,3) are the critical points.</p>
<p>
</p>
<p>Yes the horizontal line is y = 7. Thus the “top” radius would be that of the line furthest from y = 7, which is y=2x^(1/2). The “bottom” radius would be that of the line closest to y = 7, which is y = 6. Thus using your “top - bottom” rule would produce pi∫(0→9) (7-2x^(1/2))^2 - (7-6)^2 dx</p>
<p>5)</p>
<p>a) g(3) = 5+ intergral 0 - 3 g(-2) = 5 + intergral 0 - -2</p>
<p>B) 0,2,3 ?? i honestly cant remember what i even put
c) - 2, 2 , 4.5</p>
<ul>
<li>2 = minimum
2 = neither
4.5 = maximum</li>
</ul>
<p>Our teacher tought us to do (R^2 - r^2), so “big R” - “little r,” where r = distance. Thus, the longer distance is always “big R” and the shorter distance is “little r,” in which case Jersey13 is right. The distance from y = 7 to the function is longer than the distance from y = 7 to y = 6, so that one comes first.</p>
<p>Here is all of problem 5</p>
<p>A.) G(3) = 13/2 + pi, G(-2) = 5 - pi
B.) Inflection points at x = 0, 2, 3
C.) Critical points at x = 2^(1/2) and 3; x = 2^(1/2) is a relative maximum and x = 3 is neither a relative max nor a relative min</p>
<p>
That’s incorrect. You are given H(x) = G(x) - (1/2)x^2, thus H’(x) = G’(x) - x. In order to find the values at which H’(x) = 0 which are the critical points, just draw the line y = x on the graph of G’(x) and the two intersections ( at x = 2^(1/2) and 3) are the values at which H’(x) = 0.</li>
</ul>
<p>At x = 2^(1/2), the function G’(x) goes from being above the line y = x to below, meaning that H’(x) goes from positive to negative showing that x = 2^(1/2) is a relative maximum.</p>
<p>At x = 3, H’(x) remains negative on both sides of the point showing that it is not a relative extrema.</p>