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22. The base of a solid is the region enclosed by the ellipse 4x^2+y^2=1. If all plane cross sections prependicular to the x-axis are semicircles, the its volume is
(A) Pi/3
(B) Pi/6
(C) Pi/2
(D) 2Pi/3
(E) Pi/4
23. If F(x) = (Sinx)^2/(1-cosx) then F'(x) =
(A) cos x
(B) Sin x
(C) -Sin x
(D) -cos x
(E) 2cosx
24. If dy/dx= y cos x and y=3 when x=0 then y=
(A) e^(sinx)+2
(B) e^(sinx) +3
(C) sinx +3
(D) sin x +3e^x
(E) 3e^(sin x)
The volume is the integral over the range ofQuote:
Originally Posted by Nimmy
that correspond to points
on the ellipseof the areas of semi-circles of radius y. That is:
![<br />
V=\int_{-1/2}^{1/2} \frac{1}{2} \pi y^2\ dx=\int_{-1/2}^{1/2} \frac{1}{2} \pi (1-4x^2)\ dx=\frac{1}{2} \pi \left[ x - \frac{4}{3}r^3\right]_{-1/2}^{1/2}](http://latex.codecogs.com/png.latex?<br />
V=\int_{-1/2}^{1/2} \frac{1}{2} \pi y^2\ dx=\int_{-1/2}^{1/2} \frac{1}{2} \pi (1-4x^2)\ dx=\frac{1}{2} \pi \left[ x - \frac{4}{3}r^3\right]_{-1/2}^{1/2})
RonL
Quote:
Originally Posted by Nimmy
=\frac{(\sin(x))^2}{1-\cos(x)}=\frac{1-(\cos(x))^2}{1-\cos(x)}=\frac{(1-\cos(x))(1+\cos(x))}{1-\cos(x)})
Hence:
RonL
You can confirm that (E) is correct by checking the initial condition andQuote:
Originally Posted by Nimmy
differentiating and confirming that its derivative is.
Alternativly:
is a ODE of seperable type, so seperating the and variables integrating gives:
,
so:
,
or:
Imposing the initial conditionwhen
gives:
RonL