Integration (Enclosed Area & Rotation Volume)

[Latios1314]

1) Find the area enclosed by the curve x=3+cos(t) and y=4sin(t).

My tutor gave me a hint which confuses me.

A=-4 integral y dx where the upper limit is pi/2 and the lower limit is 0.

I do not understand the -4 part. why must we multiply the integral by -4?



2) Find the volume of the solid generated by revolving about the y-axis the area between the first arc of the cycloid x=t-sin(t), y=1-cos(t) and the x-axis. Use the shell method.

v=2pi integral xy dx
=2pi integral (t-sin(t)) (1-cos(t))^2 dt

I'm stuck at this point.

 

[Latios1314]

could anyone help me with these two questions?

 

[Latios1314]

anyone?=/

 

[HallsofIvy]

If x= 3+cos(t) and y= 4 sin(t), then cos(t)= x- 3 and sin(t)= y/4 so that $(x- 3)^2+ y^2/16= sin^2(t)+ cos^2(t)= 1$. The figure is an ellipse with center at (3, 0), minor semi-axis on the x-axis, of length 1, and major semi-axis parallel to the y-axis, of length 4. The area of such an ellipse is $\pi$ times the product of the two semi-axes, here, $4\pi$.

A "standard" parameterization for a circle would be "x= R cos(t), y= R sin(t)" where R is the radius and t goes around the full circle, from 0 to $2\pi$. Using different "R" for x and y (1 and 4) gives an ellipse rather than a circle. Going from 0 to $\pi/2$ gives you 1/4 of the ellipse. Because of the symmetry, multiplying that by 4 will give the area of the entire ellipse. The negative is because with x= R cos(t) (or x= cos(t)+ 3), $dx= -R sin(t)dt$.

You can also see this as a special case of Greens Theorem:
$$\int\int_D \left(\frac{\partial M}{\partial x}- \frac{\partial N}{\partial y}\right) dxdy= \oint Ndx+ Mdy$$

In particular, if we take M= 0 N= y the left side becomes $\int\int_D -1 dxdy$ which is just the negative of the area of region D. The right side becomes $\oint y dx$ where the integral is around the circumference of the region.

However, this problem has nothing to do with "volume of rotation" or even "volume".