MHB Area rotated around x-axis problem

[DeusAbscondus]

Hi folks, me again.

I have been asked to find the volume of an area rotated round the x-axis.
The area is that between two curves:
$$y=x^2$$
$$y=2x+3$$

I have calculated the area then got stuck as to what equation to plug into:
$$V=\int^b_a (f(x))^2\ dx$$

Would someone take a look at the geogebra screenshot which will follow in next post?
Thanks kindly,
D'Abs

 

[MarkFL]

I would set this up as follows:

Using the washer method:

First, find the volume of an arbitrary washer:

$\displaystyle dV=\pi((2x+3)^2-(x^2)^2)\,dx$

By sketching the graph of the region to be rotated, we see that the linear function is the outer radius and the parabola is the inner radius. The thickness of the washer is dx.

Now, sum up the washers with integration:

$\displaystyle V=\pi\int_{-1}^3 -x^4+4x^2+12x+9\,dx$

 

[DeusAbscondus]

I would set this up as follows:

Using the washer method:

First, find the volume of an arbitrary washer:

$\displaystyle dV=\pi((2x+3)^2-(x^2)^2)\,dx$

By sketching the graph of the region to be rotated, we see that the linear function is the outer radius and the parabola is the inner radius. The thickness of the washer is dx.

Now, sum up the washers with integration:

$\displaystyle V=\pi\int_{-1}^3 -x^4+4x^2+12x+9\,dx$

Thanks for this; I followed your demonstration and got the correct solution, which I've attached. It seems I could only get the solution when I set it out in the form
$$V=\pi\int^b_a\ (f(x))^2-(g(x))^2\cdot dx$$
then expanded the functions *before* integration!
Any comment(s) on why this should have made the difference for me?

Also, one very perplexing thing for me:
is it the integrand or the integral which constitutes the area ie: the term which gets squared in the formula:
$$V=\pi\int^b_a ((f(x))^2-(g(x))^2)\ dx$$
I think I've just answered my own question: by definition, it is the integrand, right?

Thanks for your great help, Mark.
Deus' Abs
PS Don't understand your point about "outer" and "inner" radii.
Can you recommend a way that I could graph this and see a 3D simulation of it?
I'm not sure if Geogebra (my main calculus learning aid) can do it.

 

[MarkFL]

You don't have to expand all of the terms in the integrand. You could write:

$\displaystyle V=\pi\int_{-1}^3(2x+3)^2-(x^2)^2\,dx=$

$\displaystyle V=\pi\int_{-1}^3 (2x+3)^2\,dx-\pi\int_{-1}^3 x^4\,dx$

Now, for the first integral, use the substitution:

$\displaystyle u=2x+3\,\therefore\,du=2\,dx$ and we have:

$\displaystyle V=\frac{\pi}{2}\int_{1}^9 u^2\,du-\pi\int_{-1}^3 x^4\,dx$

It is the integrand, including the constant out front that represents the area of the face of the washers, and the differential that represents the thickness.

Each washer has an outer radius, the circumference of which is the outside circular edge and the inner radius which pertains to the "hole" in the washer. As you can see from the integrand, we are subtracting the area of this hole from the area of the entire disk making up the washer.

As far as a program that will graph solids of revolution, I don't know of any offhand. I'm sure others here can assist with that.