Volume integral for y = (x - 2)^4, 8x - y = 16 abt x = 10

[MarkSA]

Hello,

We're doing volumes currently in Calculus.

1) Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
y = (x - 2)^4, 8x - y = 16, about the line x = 10

With most of these problems, i've started by drawing a graph of the functions and then making them into a solid to get an idea visually about what will be the radius and how to derive the area.
I'm not sure how to graph (x - 2)^4 without a graphing calculator though. I would also find the limits by doing a system of equations with both of those functions and setting them equal. But i'm not sure how I can do that with (x - 2)^4.
y = (x - 2)^4, 8x - y = 16
y = (x - 2)^4, y = 8x - 16
(x - 2)^4 = 8x - 16...? How to solve for x?

Any suggestions? How can a person find the radius without seeing or knowing the graph? Just by knowing what the curves look like in your head? Is there another way?

Thanks

 

[stapel]

I'm not sure how to graph (x - 2)^4 without a graphing calculator though.

The graph of (x - 2)[sup:360fz4gj]4[/sup:360fz4gj] is just like the graph of x[sup:360fz4gj]4[/sup:360fz4gj] (that is, a "skinnier" "parabola"), but shifted two units to the right.

...a system of equations... But i'm not sure how I can do that with (x - 2)^4.

Just use substitution:

. . . . .y = (x - 2)[sup:360fz4gj]4[/sup:360fz4gj]

. . . . .(x - 2)[sup:360fz4gj]4[/sup:360fz4gj] = 8x - 16

. . . . .x[sup:360fz4gj]4[/sup:360fz4gj] - 8x[sup:360fz4gj]3[/sup:360fz4gj] + 24x[sup:360fz4gj]2[/sup:360fz4gj] - 32x + 16 = 8x - 16

. . . . .x[sup:360fz4gj]4[/sup:360fz4gj] - 8x[sup:360fz4gj]3[/sup:360fz4gj] + 24x[sup:360fz4gj]2[/sup:360fz4gj] - 40x + 32 = 0

Use the Rational Roots Test to find potential zeroes, and then use synthetic division (and a quick graph) to find the two real zeroes. The other two zeroes are complex, so you can ignore them. :wink:

Eliz.

 

[galactus]

When using shells and revolving about a line other than an axis, you set up your integral to account for that.

To find the limits a and b, set them equal and solve for x. \(\displaystyle (x-2)^{4}=8x-16\)

\(\displaystyle x^{4}-8x^{3}+24x^{2}-40x+32=(x-4)(x-2)(x^{2}-2x+4)\)

\(\displaystyle 2{\pi}\int_{a}^{b}(x-\text{distance from origin})(f_{1}(x)-f_{2}(x))dx\)

With washers:

Set it up in terms of y. Your limits will be adjusted accordingly.

\(\displaystyle {\pi}\int_{a}^{b}\left[(f_{1}(y)-\text{distance from origin})^{2}-(f_{2}(y)-\text{distance from origin})^{2}\right]dy\)

Here's an animated graph of your rotated region.

 

[MarkSA]

Wow thanks for the reply and the graph.