How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region $y = 1 + x^{2}$ $y = 1 + x^2$ , $y = 0$ $y = 0$ , $x = 0$ $x = 0$ , $x = 2$ $x = 2$ rotated about the y-axis?

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How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region $y = 1 + x^{2}$ $y = 1 + x^2$ , $y = 0$ $y = 0$ , $x = 0$ $x = 0$ , $x = 2$ $x = 2$ rotated about the y-axis?

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Answer 1 · 2015-10-21T04:12:18

$12 π$ $12pi$

Explanation:

With a problem like this, it is always a good idea to start with a graph.

enter image source here

We are going to rotate this shape around the $y$ $y$ -axis, which will result in a cylinder with a parabolic indentation in the top. In order to evaluate this using shells, we will be integrating a bunch of ring surfaces between the $x$ $x$ -axis and the function, thus we will get an area function, $A (x)$ $A(x)$ for surfaces at each $x$ $x$ and add them all together to get the volume.

We can find our area function by stacking a bunch of circles of radius $x$ $x$ until they reach a height of $f (x)$ $f(x)$ . In other words;

$A (x) = 2 π (r a d i u s) (h e i g h t) = 2 π x (1 + x^{2})$ $A(x)=2pi (radius) (height)=2pi x(1+x^2)$

Now that we have an area function, we just need to integrate it from the $y$ $y$ -axis to $x = 2$ $x=2$ .

$\int_{0}^{2} A (x) d x$ $int_0^2 A(x) dx$

Plug in the area function.

$\int_{0}^{2} 2 π x (1 + x^{2}) d x$ $int_0^2 2pi x(1+x^2) dx$

Move the $2 π$ $2pi$ out front and simplify inside the integral.

$2 π \int_{0}^{2} x + x^{3} d x$ $2pi int_0^2 x+x^3 dx$

Solve the integral.

$2 π {(\frac{x^{2}}{2} + \frac{x^{4}}{4})}_{0}^{2}$ $2pi (x^2/2 +x^4/4)_0^2$

$2 π (\frac{2^{2}}{2} + \frac{2^{4}}{4})$ $2pi (2^2/2+2^4/4)$

$12 π$ $12pi$

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How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region $y = 1 + x^{2}$ $y = 1 + x^2$ , $y = 0$ $y = 0$ , $x = 0$ $x = 0$ , $x = 2$ $x = 2$ rotated about the y-axis?

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How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region y=1+x2y = 1 + x^2, y=0y = 0, x=0x = 0, x=2x = 2 rotated about the y-axis?

1 Answer

Explanation:

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How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region $y = 1 + x^{2}$ $y = 1 + x^2$ , $y = 0$ $y = 0$ , $x = 0$ $x = 0$ , $x = 2$ $x = 2$ rotated about the y-axis?