how do you calculate the volume of revolution of $y=\sqrt x, y=x $ around the axis $x=-2$ using the disc method. My main issue is that we work between $x=0 $ and $x=4$ but the axis is $x=-2$.
volume of revolution of $y=\sqrt x, y=x $ around the axis $x=-2$ using the disc methode
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calculus
2 Answers
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So using the disc method we have to rewrite both functions as $x=f(y)$ and then find the bounds. We see they intersect at $(0,0)$ and $(1,1)$ now because we are revolving around $x=-2$ we must shift our graphs over. So the integral we need to evaluate is $\pi\int_{0}^{1}\left ( y+2\right )^2 -\left (y^2+2\right )^2\text{d}y$ alternatively we could use the shell method (much easier) and get $2\pi\int_{0}^{1} (-2-x)\left (\sqrt x-x\right )\text{d}x$
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The general formula for using the disk method is:
$$ V = \int_{a}^{b} A(y) dy $$
We first need to think about the area of our disks (i.e. the area between the two curves that is going to be rotated about the line $ x = -2 $). This area is equal to: $ A(y) = \pi(x_2 + 2)^2 - \pi(x_1+2)^2 $, where $ x_2 $ is the function that is furthest from the line $ x = -2 $ (i.e. the outer function) and $ x_1 $ is the function that is closest to the line $ x = -2 $ (i.e. the inner function).
$$ V = \int_{a}^{b} A(y) dy = \int_{0}^{1} [\pi(x_2 + 2)^2 - \pi(x_1+2)^2] dy = \pi \int_{0}^{1} [(y+2)^2 - (y^2 + 2)] dy = 4\pi$$