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Volume of not a football

mark

Suppose we spin the region between the graph of y=\sin(x) and over the interval [0,\pi] about the y-axis We generate something that does not look at all like a football. Find the volume of that thing.

mearing

Finding the volume of not a football using shells.

y=sin(x) [0,\pi]

The height of each shell is h=sin(x) and the width of each shell is w=x.

V=2\pi \int_0^\pi xsin(x)dx

Integration by parts
f(x)=x f'(x)=1
g(x)=-cos(x) g'(x)=sin(x)

=2\pi[-xcos(x)\big|_0^\pi+\int_0^\pi cos(x) dx]= 2\pi[-xcos(x)+sin(x)]\big|_0^\pi

=2\pi(-\pi cos(\pi)+sin(\pi))= 2\pi^2

knguyen3

V=\int_0^\pi(2\pi xsin(x))dx=2\pi\int_0^\pi xsin(x)dx=2\pi((-xcos(x))|_0^\pi+\int_0^\pi cos(x)dx)\\ =2\pi(2\pi)+2\pi(sin(x)|_0^\pi)=4\pi^2

chowell1

We are finding the area of y = sin(x) with shells. to do so, we should use the formula:

V = 2\pi\int_a^bxf(x) dx

Where x is the width of each shell and f(x) is the height. Now, we plug in our expression:

V = 2\pi\int_0^{\pi}xsin(x)\ dx

from here, we can integrate by parts:

f(x) = x \ \ \ f'(x) = 1\\ g(x) = -cos(x) \ \ \ g'(x) = sin(x)

=2\pi(-xcos(x)|^{\pi}_0 \ +\ \int^{\pi}_0-cos(x)dx \ = 2\pi[-xcos(x)-sin(x)]|_0^{\pi}

=2\pi(-\pi cos(\pi) \ + \ sin(\pi)) = 2\pi(\pi - 0) = 2\pi^2