# Area Under a Curve

• Sep 2nd 2009, 05:03 AM
Lukybear
Area Under a Curve
Also if some1 can help me in this question that would be great.

Find the equation of the tangent to the parabola y=2x^2 at (1,2). Calculate its point of intersection with the x-axis and the volume of the solid formed when the area between the parabola, the tangent line and the xaxis is reveloved about the xaxis.

Especially the bit about forming a solid of revolution.

The tangent, has equn of
y=4x-2
THe pt of intersection is (1/2,0)

But the area formed i found to be 2/15 pi, which is different to the answer. Much help would be appreciated.
• Sep 2nd 2009, 05:22 AM
Calculus26
I believe 2pi/15 is correct

Draw a diagram and you'll see you need 2 integrals

between x = 0 and 1/2 you have disks

V= (pi)integral(4x^4dx) = pi/40

between x =1/2 and 1 you have washers

V = pi integral (4x^4 - (4x-2)^2)dx) =13pi/120

Adding the 2 you get 16pi/120 = 2pi/15
• Sep 2nd 2009, 05:35 AM
Calculus26
See attachment for diagram and set up
• Sep 3rd 2009, 12:11 AM
Lukybear
Quote:

Originally Posted by Calculus26
See attachment for diagram and set up

Thanks SO much. Especially for the diagram.