Calculus Problems


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To do the first problem, just sketch out the graph and then find the area of the shaded area in the first, second, and fourth quadrants. First quadrant is just a triangle, but the second and fourth require integrals.

For the next part, just plug 0 in for x and y to get the intersection points, then plug in a few more numbers to get a rough idea of how it should look. For finding the volume of it after rotating it around the y-axis, you'll need a formula that I don't remember off the top of my head.

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Well, you should know how to graph these.

You know -y^2 is a paraboloa that is sideways facing to the left. And the other is just a line.

First you take the basic shape that you know, then just plug in points for x to get a y, and do that like 3 times for each and you have a basic shape.

As for finding the area, for a dy integral it is right minus left and for a dx it is top minus bottom... then integrate at the intersection points of each graph.

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Er...could you please post the questions? I don't have adobe reader. But from what I see, y^2, -y^2, y^2 +/- c are elementary graphs....parabolas. For example -y^2 is a parabola opening towards the left, and touching the y-axis at origin.

EDIT: Nevermind, I just used these neat pdf converters.

Q2. We try to reduce the equation into (x - constant)^2 = (y - constant)^2 + constant, where each constant may be different.

The given equation then becomes x^2 + 1 = (y-1)^2

This parabola has vertex as (-1, 1)

The other equation is a straight line.

The area sketched would be like the one attached. To compute it, just use integrals.

post-69025-1193609718_thumb.jpg

Edited by gaurav
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The second question will be done like this:

Volume swept when revolved by a small angle dθ = 2*(2*1*π*dθ)

Volume of the entire solid = integral (4*π*dθ) ***The limits are 0 to π

= 4*π*π

Voila!

post-69025-1193610064_thumb.jpg

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Wow, I did'n know Neowin provided math help :o

I did that last semester at uni, can't really explain it much more than m0fo or gaurav though :/

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Dammit I think I screwed up both questions....lol

So here's the correct way (I promise its right this time!):

The ellipse one seems wrong, so I'll do it:

In the integration, he took π*y^2*dx. This is when ellipse is revolved around X-axis. The question says Y-axis, so you take integral (π*x^2*dy) with the limits -2 -> 2. Solve it and the answer comes to be (8 π)/3

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