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[Sunyata]

http://www.mathhelpforum.com/math-help/JSREPL96586:;<!-- google_ad_section_end -->

For (a), is the answer:
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For (b), do I just say the integral gives the area under the semi-circle and above the x-axis between x = 0 and x = b ???

 

[addikaye03]

http://www.mathhelpforum.com/math-help/JSREPL96586:;<!-- google_ad_section_end -->

For (a), is the answer:
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For (b), do I just say the integral gives the area under the semi-circle and above the x-axis between x = 0 and x = b ???

a) Let x=asinu [same thing, jsut rearranged]

dx/du=acosu ---> dx=acosu du

Change limits: when x=b, u=sin^-1(b/a)

x=a, a=asinu therefore u=pi/2

int. (sin^-1(b/a)-->pi/2) rt (a^2-a^2sin^2@) . acosu du

int. (sin^-1(b/a)-->pi/2) rt [a^2(cos^2@] . acosu du

int. (sin^-1(b/a)-->pi/2) a^2cos^2@ du where a is a constant

Consider cos2@=2cos^2@-1 so cos^2@=1/2(cos2@+1)

int. (sin^-1(b/a)-->pi/2) a^2/2 (cos2@+1)

so then just integrate cos2@+1 and use limits, and expand the a^2/2 and you'll get the answer.

b) So yeah, it's a semi circle, above the x-axis, you could say:

when y=0, x^2-a^2=0 hence x=+-a, these are the points at which it crosses the x-axis

when x=0, y=rt(a^2)= a so crosses the y-axis at a.

Then just sketch it.