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Calculus/double integral

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Expert: - 7/22/2011


Question
double integral
double integral
Evaluate"integral 0 to 2a integra 0 to sq.rt(2ax-x^2)  '(x^2)'dx dy"

Answer
First, it should be "dy dx" not "dx dy"...

So the inner integral is: integral of x^2 dy from y=0 to sqrt(2ax-x^2)
which is x^2*y from y=0 to y=sqrt(2ax-x^2)
= x^2*sqrt(2ax-x^2)-0

So the outer integral is now
integral x^2*sqrt(2ax-x^2) dx from x=0 to x=2a
completing the square on 2ax-x^2 gives a^2-((x-a)^2

So, we get integral x^2*sqrt(a^2-(x-a)^2) dx, x from 0 to 2a
now make the trig subs. x-a=a*sin(t), so x=a+a*sin(t), dx=s*cos(t) dt

substituting into the integral we get
integral a^4*(1+sin(t))^2*a*cos(t)*a*cos(t) dt, t from -pi/2 to pi/2

after some trig identities we get
a^4*5*pi/8

OK?

Abe

Abe Mantell

Expertise

Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!

Experience

Over 15 years teaching at the college level.

Organizations
NCTM, NYSMATYC, AMATYC, MAA, NYSUT, AFT.

Education/Credentials
B.S. in Mathematics from Rensselaer Polytechnic Institute
M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook

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