Rhonda S.

# how do u deal with the 1 in integral(from 0 to 3) (1+sqrt(9-x^2)dx

how to solve with extra addend 1 in the integral?

## 4 Answers By Expert Tutors

By: Biology, Chemistry, Math, and K-12 Tutor Michael B.

It may not be apparent to the OP, but ?v(9-x2)dx isn't a simple one, as you can't use u-substitution.  There's two ways to handle this.  First is to realize that this is the equation of a circle with radius=3, and thus the integral is asking for the area in the first quadrant.  So you can just calculate area of the circle and divide it by 4 - that's the "graphical" method.

To actually integrate this, the simplest method is to convert it to polar form:

y = v(9-x2)
y2 = 9 - x2
x2 + y2 = 9
r2 = 9
r = 3
? (1/2)(3)2 d?   (from 0 -> pi/2)

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11/15/12 Michael B.

Wow this editor doesn't work well at all...    It messed up a bunch of characters.  That last line should read "integral of (1/2)32 times d-theta"

Test: ? &theta; &#952; T

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11/15/12 Certified High School AP Calculus and Physics Teacher Masters of Education Graduate with Mathematics Expertise Michael B.

Roman, not quite correct....   Your main error is a simple one - the limits are 0..3, not -3..3.  The answer should only be 3+9pi/4.

Second, to be clear, the original equation is NOT a semi-circle, it is a circlular-curve shifted up by 1.  That is, if you graph sqrt(9-x2) then you don't get a circle, you only get the positive half of the circle (above the x-axis).  This graph is then shifted up by 1.  And then, you only consider from 0..3.  The area we are considering is a "quarter-circle or radius 3 sitting on top of a 3x1 rectangle in the first quadrant".

Otherwise, your math is correct, although somewhat over-complicated and prone to mistakes.  A cleaner method is to convert to polar coordinates, as I said in my other comment:

A = 3 + Integral from 0..3 of 1/2 32 dTheta

(apologies for the use of words - I've discovered that the symbols aren't rendered correctly in comments.)  The '3' comes from the area of the rectangle as a result of splitting the integral into f(x)=1 and g(x)=sqrt(9-x2) and the Integral from 0..3 1/2(32)dTheta is the area of the unshifted circle in the first quadrant.  This gives 3+9pi/4, the correct answer.

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11/15/12 