Find the volume of the solid

[MilfordStanton]

I have a problem like this but I don't know how to solve it. Help me

Find the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and y = 2x - x^2 about the x-axis using integration.

Thank you very much in advance.

[janusz55]

Theorem
Let \( f \) and \( g \) be continuous functions and nonnegative on \( [a, b], \) and suppouse that \( f(x) \geq g(x) \) for all \( x \) in interval \( [a,b]. \) Let \(\mathcal {R} \) be the region that is bounded above by \( y = f(x) \) below by \( y = g(x) \), and on the side \( x= a \) and \( x = b, \) then volume of the solid of revolution that is generated by revolving the region \(\mathcal{R} \) about the \( x \)-axis:

\( |V| = \pi \int_{a}^{b} ([f(x)]^2 - [g(x)]^2)dx \)


We find the limits of the integral:

\( \begin{cases} y = 2x -x^2 \\ y = x^2 \end{cases} \)

\( 2x -x^2 = x^2, \ \ 2x^2 -2x = 0, \ \ 2x(x-1) = 0, \ \ x_{1} = 0 = a , \ \ x_{2} = 1 = b.\)

Region \( \mathcal{R}\)

\( \mathcal{R} = \{(x,y) \in \rr^2: f(x) = 2x -x^2 \geq x^2 = g(x), \ \ f(x) - g(x) = (2x -x^2) - x^2, \ \ x\in [ 0, 1] \}.\)

Hence

\( |V| = \pi \int_{0}^{1} ([2x -x^2]^2 - [x^2]^2)dx = \pi \int_{0}^{1}(4x^2 -4x^3 +x^4 -x^4)dx = \pi\int_{0}^{1}(4x^2 - 4x^3)dx = 4\pi\int_{0}^{1} (x^2 - x^3)dx = 4\pi \left[ \frac{x^3}{3} - \frac{x^4}{4}\right]_{0}^{1} = 4\pi\left( \frac{1}{3}- \frac{1}{4}\right) = \)

\(= 4\pi\cdot \frac{1}{12} = \frac{1}{3}\pi.\)

[carry2693a]

With these problems, the absolute best thing to do is to draw a picture first. Visualizing helps understand your next step and how to set up your equations. Shell method is nothing but taking a line of height h, at radius r, and revolving it around whatever axis. Then finding the surface area of it, and doing that over and over to add up to the volume of the entire object. So by drawing a picture, it's easier to figure out what your r is and what equation you need for h.....

[janusz55]

I am yet to make a drawing?
You're now exaggerating with your wisdom.