Area, Volume, and Integral

[calculusqueen]

Let f and g be the functions given by f(x)=2x(1-x) and g(x)=3(x-1)(square root of x) for 0 is less than or equal to x is less than or equal to 1. The graphs are shown in the figure above.

a) Find the area of the shaded region enclosed by the graphs of f and g.

b) Find the volume of the solid generated when the shaded region enclosed by the graphs of f and g is revolved about the horizontal line y=2.

c) Let h be the function given by h(x)=kx(1-x) for 0 is less than or equal to x is less than or equal to 1. For each k is greater than 0, the region enclosed by the graphs of h and g is the base of a solid with square cross sections perpendicular to the x-axis. There is a value of k for which the colume of this solid is equal to 15. Write, but do not solve, an equations involving an integral expression that could be used to find the value of k.

 

[Deleted member 4993]

Let f and g be the functions given by f(x)=2x(1-x) and g(x)=3(x-1)(square root of x) for 0 is less than or equal to x is less than or equal to 1. The graphs are shown in the figure above.

a) Find the area of the shaded region enclosed by the graphs of f and g.

b) Find the volume of the solid generated when the shaded region enclosed by the graphs of f and g is revolved about the horizontal line y=2.

c) Let h be the function given by h(x)=kx(1-x) for 0 is less than or equal to x is less than or equal to 1. For each k is greater than 0, the region enclosed by the graphs of h and g is the base of a solid with square cross sections perpendicular to the x-axis. There is a value of k for which the colume of this solid is equal to 15. Write, but do not solve, an equations involving an integral expression that could be used to find the value of k.

These are ole standard AP calculus questions. Very straight forward no complications involved....

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[BigGlenntheHeavy]

\(\displaystyle V = \pi \int_{0}^{1}[(3x^{3/2}-3x^{1/2})-2]^{2}-[(2x-2x^{2})-2]^{2})]dx = \frac{103\pi}{20}.\)

I re-edited this, you are right galactus, I was wrong, sorry.

 

[galactus]

I ran it through Maple to check. It gave me \(\displaystyle \frac{103\pi}{20}\), too. I will post the graph later this afternoon.

 

[Deleted member 4993]

Galactus - I think you missed the center of rotation (axis)

b) Find the volume of the solid generated when the shaded region enclosed by the graphs of f and g is revolved about the horizontal line y=2.

 

[BigGlenntheHeavy]

Galactus, you are right, I am wrong.

You original volumne is correct, I stand corrected, sorry.

 

[galactus]

Thanks, Glenn.

I will post the graph. It is cool because it looks like a cored apple.