find the volume of the surface of revolution created by revolving the region bounded by the equations y=4-x.^2 and y=0 about the line x=4 in matlab

To find the volume of the surface of revolution, we need to integrate the surface area of infinitesimal cylindrical shells along the axis of rotation.

The equation of the curve is given by y = 4 - x^2. So, we can express x in terms of y as x = sqrt(4-y).

The axis of rotation is x = 4. So, the distance of the curve from the axis is r = 4 - x = 4 - sqrt(4-y) = 2 + y/(2*sqrt(4-y)).

The surface area of an infinitesimal cylindrical shell of radius r and height ds is given by dA = 2pir*ds.

To find ds, we make use of the fact that the curve is obtained by rotating the function y = 4 - x^2 about the x-axis. So, ds is given by ds = sqrt(1 + (dy/dx)^2)*dx.

We already have x in terms of y. Differentiating with respect to y, we get dx/dy = -y/sqrt(4-y). Therefore,

ds = sqrt(1 + (dy/dx)^2)dx = sqrt(1 + (1/(4-x))^2)(-2x)dy = sqrt(1 + (1/(4-(4-y)))^2)(-2sqrt(4-y))dy = sqrt(1 + (1/y)^2)(-2sqrt(4-y))dy

The volume of the surface of revolution is given by the integral of the surface area of the cylindrical shells dV = ∫dA = ∫2pirds = ∫2pi*(2+y/(2sqrt(4-y)))sqrt(1 + (1/y)^2)(-2sqrt(4-y))dy.

To evaluate this integral in Matlab, we use the following code:

main.m
syms y;
r = 2 + y/(2*sqrt(4-y));
ds = sqrt(1 + (1/y)^2)*(-2*sqrt(4-y));
dA = 2*pi*r*ds;
V = int(dA, y, 0, 4);
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The answer will be in cubic units.