Lecture Notes 4-9-18 though 4-13-18

LectureNotes 4-9-18

LectureNotes 4-11-18

LectureNotes 4-13-18

13.4#4

I got the integral dx (x^2) +  (8x) dy= 2x + 0

(7,0) (7+y,y)  2x +0

But it doesn't seem to work

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OK, to use Green's theorem you need the formula ∬ ∂F_2/∂x - ∂F_1/∂y dA = ∫_C F_1dx + F_2dy, where the line integral is counter clockwise, while you seem to have computed ∂F_1/∂x + ∂F_2/∂y. Also, note that the line integral in question is computed clockwise, which will change the sign of the left side the above equation. 

13.4#3

I'm not sure what I got wrong.  I came out with the integral 
(3sqrt(2)/2),(0)      (sqrt(9-y^2)^2), (y)    4-2 dy dx

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Well it looks like you did two things wrong. The first and most important is that ∂F_1/∂y = -2, so you should have had 4-(-2) =6, not 4-2=2. The second might be that your integral should have been dxdy not dydx.  Also, a) the integrand is a constant and so polar integration might be easier, and b) the area of a pie wedge is (1/2)(Δθ)r^2, which is to say (9π/8) you could use that instead of doing an integral

LectureNotes from 3/19/18 through 4/6/18

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make $40

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Estimated Grade so far

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12.3#15

I don't understand this problem. I've tried multiplying the base and hieght and tried to do this:
I (pi 0) I (8 0 )(x=rcosA - r^2)r dr dz dA

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well, its hard to figure out what you mean but I guess maybe you mean A=θ, and I guess you're doing a triple integral, which could work but is excessive for a homework problem dealing with polar integrals (and wrong as you've implemented here).

From the way the plane z=x cuts the xy-plane at the y-axis,  and from the fact that the region is above the xy-plane, you have a base that's a semidisk in the x≥0 part of the xy-plane and a height (z) that's equal to x above that, which gives the integrals

∫_{-π/2}^{π/2} ∫_0^8 r cos(θ) r dr dθ,

so you kind of have the integrand except for that -r^2) part.  The limits of integration in r are fine, but the limits of integration in θ would be for the plane z=y instead of z=x.