<p>lol im pretty desperate so please help me.</p>
<p>ok</p>
<p>Directions: I need to derive a formula for the Volume of the solid by slicing:
1) A right-Circular Cone of altitude h units and base radius of a units.</p>
<p>Next one you dont have to use the directions.</p>
<p>2) The base of a solid is the region enclosed by an ellipse having the equation 3x^2 + y^2 = 6. Find the volume of the solid if all plane sections perpendicular to the x axis are squares. </p>
<p>If you can tell me how to do those two, i can do the rest and if you do help me you would the greatest person alive for the next month in My world.</p>
<p>Bump Bump Bump Bumpety Bump Bump</p>
<p>AHHHHHHHHHHHHHHHHHHHHHHHH can you guys help me?</p>
<ol>
<li>so youre going to use the formula for area of a circle and really all youre doing is riemann sums, in this case, the sums of the volumes of disks with height da and face area a. see what im saying? the area for a circle of yours would be pi(a^2) then times the height, which in this case is da. since you're going to certain height, or bound, id suppose you want to take h as the upper bound and say:</li>
</ol>
<p>Pi* (Integral[0,h] of a^2 da)</p>
<h1>2. this is hard to explain without a cool picture but make the equation y = sqrt. (6-3x^2) and itll give you half of the ellipse. half of the base of one of those squares is y, 2y is the whole base. so the area of the faces of one of those squares is 2y^2, then for volume, you pull in the height of each cube thingy and call it dx. 2y in terms of x is 2(sqrt. 6-3x^2)^2, the squares cancel and you get 12-6x^2. then do the integral from i think -sqrt2 to sqrt 2 and youll get the answer:</h1>
<p>Integral[-sqrt.2, sqrt.2] of 12-6x^2 dx</p>
<p>or about 22.627</p>
<p>someone correct me if these are wrong. volumes and cross sections are my weakest area of calculus so help me out if i just suck. lol</p>
<p>and if i am right, i look forward to being the greatest person alive in your world for the next month. now, want to do my history homework? lol</p>