integral volume question (1 Viewer)

[member 6003]

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 x axis is revolved about the axis.
got a different answer, its apparently 2pi/3 for the volume

 

[cossine]

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 x axis is revolved about the axis.
got a different answer, its apparently 2pi/3 for the volume

show your working out

 

[member 6003]

show your working out

 

[cossine]

Well done. You are extremely close.

What you have sort of done is said,

(a+b)^2 = a^2 + b^2 # However this is not true since (a+b)^2 = a^2 + b^2 + 2ab.


So what you should have as your integrand i.e. what your integrating is (2x^2 - 4x + 2)^2 # the square is because area circle is pi*r^2.

(2x^2 - 4x + 2)(2x^2 - 4x + 2)

https://www.emathhelp.net/algebra-c...+2&q=2x^2+-+4x++2&action=multiply+polynomials
Once we expand this we get:



All that is left now is is integrate with respect x with bounds from 0 to 1, and don't forget to times by pi.