- #1
Lil_Margie
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Homework Statement
So I have three questions. The first one is the double integration of x+2 from y=0 to y=sqrt(9^2-x^2). The second question is the double integration of sqrt(r^2-x^2-y^2) where the domain is in the circle of radius R and origin 0. And the last question is the double integration of ax^3+by^3+sqrt(a^2-x^2), where xis from -a to a and y is from -b to b. Please no polar coordinates!
Homework Equations
The Attempt at a Solution
1. I drew a semicircle where the radius was 3. The x is therefore from -3 to 3. And the y can be simplified to be from 0 to 3*pi, since the outer border of the semicircle is half the circumference, which is 2*pi*r. So, (2*pi*3)/2 is 3*pi. Therefore x is from -3 to 3 and y is from 0 to 3*pi. After I integrated, first with respect to y, then to x, I got 36*pi. But the textbook answer is 9*pi. Why?
2. I don't really know where to start. I just know that I'll have z^2+x^2+y^2=r^2 and that on the xy plane z=0 so I have x^2+y^2=r^2 which is one of the boundaries. So I can have the boundaries 0≤y≤sqrt(r^2-x^2) and -r≤x≤r, say if I'm only focusing on the semicircle which lies on the xy plane. But after that I don't know what to do!
3. I first simplified the formula by noticing the sqrt part can be drawn as a semicircle, with radius a. This means that the circumference of that semicircle will be 2*pi. So I can put in 2*pi and get rid of the sqrt equation. Then by integrating first with respect to y, then to x, and using the boundaries I mentioned, I get 4*a^2*b*pi. I don't know how to get rid of the 4!
Thanks.