Application of Integrals

[tkvictim]

The directions read:
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

I can't seem to sketch the graph for most of my problems, which is probably one of the most important things to get down.

y=(1/4)x[sup:s44j6say]2[/sup:s44j6say], y=5-x[sup:s44j6say]2[/sup:s44j6say]; about the x-axis.

My graph looks like a snowman... Can someone guide me on to where to begin?

 

[arthur ohlsten]

y=[1/4] x^2
y=5-x^2

second order equation a conic section

1) y=[1/4] x^2 a parabola vertex at 0,0 open up

2) y=5-x^2
y-5 = -x^2 a parabola, vertex at 0,5 open down

you should be able to sketch both parabola's

Arthur

 

[BigGlenntheHeavy]

\(\displaystyle y \ = \ \frac{x^{2}}{4}, \ y \ = \ 5-x^{2}\)

\(\displaystyle V_{disc} \ = \ 2\pi\int_{0}^{2}\bigg[(5-x^{2})^{2}-\bigg(\frac{x^{2}}{4}\bigg)^{2}\bigg]dx \ = \ \frac{176\pi}{3}\)

\(\displaystyle See \ graph.\)

[attachment=0:2mtm7gar]xyz.jpg[/attachment:2mtm7gar]

 

[Deleted member 4993]

y=(1/4)x[sup:2q1mh01w]2[/sup:2q1mh01w], y=5-x[sup:2q1mh01w]2[/sup:2q1mh01w]; about the x-axis.

My graph looks like a snowman... Can someone guide me on to where to begin?

It does look a bit like snow-man (more like weebles).

 

[galactus]

Just for fun, here is the shells method. It is a little more daunting than the washers, though.

\(\displaystyle 4{\pi}\left[\int_{0}^{1}y\cdot 2\sqrt{y}+\int_{1}^{5}y\sqrt{5-y}\right]dy\)

Here is an animated graph showing the disks:

 

[tkvictim]

Thanks for the visuals!

Is the problem evaluated from [0,2] or [0,5]? What are the upper and lower bounds?

 

[tkvictim]

y=(1/4)x[sup:gowlygwe]2[/sup:gowlygwe], y=5-x[sup:gowlygwe]2[/sup:gowlygwe]; about the x-axis.

My graph looks like a snowman... Can someone guide me on to where to begin?

It does look a bit like snow-man (more like weebles).

Hahahahaha you are spot on!

 

[tkvictim]

Here is another problem I'm having a little trouble on:

Find the numbers b such that the average value of f(x) = 2 + 6x - 3x[sup:2nk06wen]2[/sup:2nk06wen] on the interval [0,b] is equal to 3.

My teacher never exposed us to a problem where we had to find the value of the upper limit, so I'm not sure how to approach this. I have the avg value function set up, but it doesn't help much.

 

[BigGlenntheHeavy]

\(\displaystyle \frac{1}{b}\int_{0}^{b}[2+6x-3x^{2}]dx \ = \ 3, \ now \ solve \ for \ b.\)