I need calculus help...

[Giggles The Panda]

The tangent lines to the parabola y=ax^2+b at the point (2,3) and (-2,3) contain the origin.

(a) Find the value of a and b.
(b) Find the area of the region R enclosed by the parabola and the two tangent lines.
(c) Set up and evaluate an integral expression in one variable which gives the volume when the region R, described in part (b), is revolved about the y-axis.

Please show your work and maybe give a brief explanation of what you did.

[Mathematics]

The tangent lines to the parabola y=ax^2+b at the point (2,3) and (-2,3) contain the origin.

(a) Find the value of a and b.
(b) Find the area of the region R enclosed by the parabola and the two tangent lines.
(c) Set up and evaluate an integral expression in one variable which gives the volume when the region R, described in part (b), is revolved about the y-axis.

Please show your work and maybe give a brief explanation of what you did.

(a) Find the value of a and b.

The form of both lines is:

y=mx,

as there is no constant (the lines pass throught the origin). The gradients of the lines are then given by dy/dx:

y=\frac{3}{2}x

and

y=-\frac{3}{2}x.

We now know what the gradient of the curve must be at these points, so we can differentiate and find the constant a:

y=ax^2+b,

\frac{\partial y}{\partial x}=2ax.

Substituting in the gradients and the values of x:

2 a (2)=3/2

2 a (-2)=-3/2

a=3/8.

So we have:

y=\frac{3}{8}x^2+b.

From the lines we also know the values of y at x=2 and x=-2, so we can find b:

3=\frac{3}{8}2^2+b,

b=3/2.

(b) Find the area of the region R enclosed by the parabola and the two tangent lines.

The basic stratergy is to integrate between 0 and x=2 (where the line touches the curve), take away the area between the line and the x-axis (just a triangle) and double the answer, as the curves are symmetric in the y-axis.

\int _0^2 \left(\frac{3}{8}x{}^2+\frac{3}{2}\right) dx=4

(Click "Show steps" for details: http://www81.wolframalpha.com/input/?i=Integrate+3%2F8x^2%2B3%2F2)

Area between line and x-axis:

\frac{1}{2}\left(base \times height \right) = \frac{1}{2}(2)(3)=3

Area under curve - area of triange = 1

So total area enclosed by the curve and the two lines = 2*1 = 2.

[Mathematics]

(c) Set up and evaluate an integral expression in one variable which gives the volume when the region R, described in part (b), is revolved about the y-axis.

Volume of revolution when a function is rotated around the y-axis:

V=\int \pi x^2 dy

Notice that the function has been changed so it is in terms of y i.e. instead of writing y=... you are re-arranging and writing x=... (function of y).

If you imagine the volume swept out by rotating the line y=3/2x around the y-axis between y=0 and y=3 you will have a cone of volume:

V_1=\int_{0}^3 \pi \left(\frac{2}{3}y\right)^2 dy=4\pi

You can then simply subtract the volume swept out by the quadratic as it rotates. The quadratic doesn't touch the x-axis; at x=0 y=3/2. So you need to integrate between y=3/2 and y=3.

V_2=\int_{3/2}^3 \pi \left(\frac{8}{3} \left(y-\frac{3}{2}\right) \right)dy=3\pi


V_R=V_1-V_2=\pi