Calculus BC · Unit 8
Square the radius for a disc. For a washer, square the outer radius and the inner radius, then subtract.
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Disc
Revolving a region that reaches the axis:
\(V=\pi\int_a^b R(x)^2\,dx\)
\(R\) is the distance from the curve to the axis of rotation.
Washer
A hole, or a gap between the region and the axis:
\(V=\pi\int_a^b \left(R_{\text{out}}^2-R_{\text{in}}^2\right)\,dx\)
The axis decides the radius
Around the x-axis, a function \(y=f(x)\) has radius \(|f(x)|\). Around \(y=-1\), the radius is the distance \(|f(x)-(-1)|\).
dy for a revolution around the y-axis
Write \(x\) as a function of \(y\) when you use washers perpendicular to the y-axis.
One radius, and the region meets the axis.
Radius \(R=\sqrt{x}\), so \(R^2=x\).
\(V=\pi\int_0^4 x\,dx\)
\(\pi\left[\dfrac{x^2}{2}\right]_0^4\)
\(8\pi\)
Answer: \(8\pi\)
\(V=\pi\int_0^3 x^2\,dx\)
\(\pi\cdot 9\)
\(9\pi\)
Answer: \(9\pi\)
This is a cylinder of radius 2 and height 5.
\(\pi\cdot 4\cdot 5\)
\(20\pi\)
Answer: \(20\pi\)
\(V=\pi\int_0^1 x^4\,dx\)
\(\dfrac{\pi}{5}\)
Answer: \(\dfrac{\pi}{5}\)
Square the radius \(\sqrt{x}\) to get \(x\).
Multiply by \(\pi\).
Answer: A. \(\pi\int_0^4 x\,dx\)
Radius \(x=y^2\).
\(V=\pi\int_0^2 y^4\,dy\)
\(\pi\left[\dfrac{y^5}{5}\right]_0^2\)
\(\dfrac{32\pi}{5}\)
Answer: \(\dfrac{32\pi}{5}\)
A washer: outer radius and inner radius.
On \([0,1]\), the line is farther from the x-axis than the parabola.
\(R_{\text{out}}=x\) and \(R_{\text{in}}=x^2\).
\(V=\pi\int_0^1 (x^2-x^4)\,dx\)
\(\pi\left[\dfrac{x^3}{3}-\dfrac{x^5}{5}\right]_0^1\)
\(\pi\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(\dfrac{2\pi}{15}\)
Answer: \(\dfrac{2\pi}{15}\)
Outer radius, from \(y=-1\) up to \(y=x\): \(x+1\).
Inner radius, from \(y=-1\) up to \(y=0\): \(1\).
\(V=\pi\int_0^2 \left((x+1)^2-1\right)\,dx\)
\((x+1)^2-1=x^2+2x\)
\(\pi\left[\dfrac{x^3}{3}+x^2\right]_0^2\)
\(\pi\left(\dfrac{8}{3}+4\right)\)
\(\dfrac{20\pi}{3}\)
Answer: \(\dfrac{20\pi}{3}\)
\(x=y^2\) and \(y\) runs from 0 to 2. The outer edge is \(x=4\).
\(V=\pi\int_0^2 (4^2-(y^2)^2)\,dy\)
\(\pi\int_0^2 (16-y^4)\,dy\)
\(\pi\left[16y-\dfrac{y^5}{5}\right]_0^2\)
\(\pi\left(32-\dfrac{32}{5}\right)\)
\(\dfrac{128\pi}{5}\)
Answer: \(\dfrac{128\pi}{5}\)
Outer radius 4, inner radius \(x^2\), from \(-2\) to \(2\).
\(V=\pi\int_{-2}^{2}(16-x^4)\,dx\)
\(2\pi\int_0^2 (16-x^4)\,dx\)
\(2\pi\left[16x-\dfrac{x^5}{5}\right]_0^2\)
\(2\pi\left(32-\dfrac{32}{5}\right)\)
\(\dfrac{256\pi}{5}\)
Answer: \(\dfrac{256\pi}{5}\)
\(V=\pi\int_0^4 (9-1)\,dx\)
\(\pi\cdot 8\cdot 4\)
\(32\pi\)
Answer: \(32\pi\)
Both curves are below \(y=2\).
The curve farther from \(y=2\) is the lower one, \(y=x^2\).
Outer radius \(2-x^2\).
Inner radius \(2-x\).
Answer: outer \(2-x^2\), inner \(2-x\)
Finish the washer around a line that is not an axis.
\(V=\pi\int_0^1 \left((2-x^2)^2-(2-x)^2\right)\,dx\)
\((2-x^2)^2=4-4x^2+x^4\)
\((2-x)^2=4-4x+x^2\)
Difference: \(x^4-5x^2+4x\).
\(\pi\left[\dfrac{x^5}{5}-\dfrac{5x^3}{3}+2x^2\right]_0^1\)
\(\pi\left(\dfrac{1}{5}-\dfrac{5}{3}+2\right)\)
\(\pi\left(\dfrac{3-25+30}{15}\right)\)
\(\dfrac{8\pi}{15}\)
Answer: \(\dfrac{8\pi}{15}\)
\(y\) runs from 0 to 2. At height \(y\), the region runs from \(x=y\) to \(x=2\).
Distance to \(x=3\): outer radius \(3-y\), inner radius \(3-2=1\).
\(V=\pi\int_0^2 \left((3-y)^2-1\right)\,dy\)
\((3-y)^2-1=8-6y+y^2\)
\(\pi\left[8y-3y^2+\dfrac{y^3}{3}\right]_0^2\)
\(\pi\left(16-12+\dfrac{8}{3}\right)\)
\(\dfrac{20\pi}{3}\)
Answer: \(\dfrac{20\pi}{3}\)
\(V=\pi\int_0^1 e^{2x}\,dx\)
\(\pi\left[\dfrac{1}{2}e^{2x}\right]_0^1\)
\(\dfrac{\pi}{2}(e^2-1)\)
Answer: \(\dfrac{\pi}{2}(e^2-1)\)
Right edge \(x=e\), left edge \(x=e^{y}\).
\(V=\pi\int_0^1 \left(e^2-(e^{y})^2\right)\,dy\)
\(\pi\int_0^1 (e^2-e^{2y})\,dy\)
\(\pi\left[e^2 y-\dfrac{1}{2}e^{2y}\right]_0^1\)
\(\pi\left(\left(e^2-\dfrac{e^2}{2}\right)-\left(0-\dfrac{1}{2}\right)\right)\)
\(\pi\left(\dfrac{e^2}{2}+\dfrac{1}{2}\right)\)
\(\dfrac{\pi}{2}(e^2+1)\)
Answer: \(\dfrac{\pi}{2}(e^2+1)\)
\(V=\pi\int_1^6 (25-4)\,dx\)
\(21\pi\cdot 5\)
\(105\pi\)
Answer: \(105\pi\)
That integral squares the gap between the curves.
A washer squares each radius separately, then subtracts.
\(\pi\int_0^1 (x^2-x^4)\,dx\) is the volume.
Squaring the difference misses the hole’s position relative to the axis.
Answer: square each radius, then subtract
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