How do you find the volume of the bounded region if $y = sin x$ $y = sinx$ , $y = 0$ $y = 0$ from $x = \frac{π}{4}$ $x = pi/4$ , $x = 3 \frac{π}{4}$ $x = 3pi/4$ , revolved around the y-axis?

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Answer 1 · 2015-06-19T00:17:26

Use the shell method.

Representative shell has volume: $2 π r h \cdot thickness$ $2pi r h * "thickness"$

$r = x$ $r = x$ , $h = sin x$ $h = sinx$ and $thickness = d x$ $"thickness" = dx$

So the volume of the solid is given by:

$V = \int_{\frac{π}{4}}^{\frac{3 π}{4}} 2 π x sin x d x$ $V = int_(pi/4)^((3pi)/4) 2pi x sinx dx$

$V = 2 π (sin x - x cos x)]_{\frac{π}{4}}^{\frac{3 π}{4}}$ $V =2pi( sinx-xcosx)]_(pi/4)^((3pi)/4$

Then use trigonometry and arithmetic to get:

$V = 2 π \cdot \frac{π \sqrt{2}}{2} = π^{2} \sqrt{2}$ $V =2pi* (pi sqrt2)/2 = pi^2 sqrt2$

How do you find the volume of the bounded region if y=sinxy = sinx, y=0y = 0 from x=π4x = pi/4, x=3π4x = 3pi/4, revolved around the y-axis?