Estimate Grades so far

(Double click on the image to enlarge)

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

*******************

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.

12.3#9

I'm unsure what a and b are equal to. I'm pretty sure they should be a = -pi/2 and b = pi/2



**************************
Well no, but you could think that if you didn't understand the geometry.  Did you follow the instructions and graph those curves?  Because that would have been very instructive, since r=1/(3 cos(θ)), which is the same as  r cos(θ)=1/3,  which is the same as x=1/3, which is a vertical straight line at x=1/3.  The constraint  -π/2≤θ≤π/2 means that you've got the whole line from y = -∞ to y = ∞.  You only get the part of the line inside the curve r=1,  which is the unit circle,  which has the equation x^2 + y^2 = 1, and when you substitute x=1/3,  you can get y^2 = 2/3 or y = ± √2/√3, which are a long way from ±∞.  Now that you have an x and a y, you should be able to use them to find your limits in θ.

12.3#10

The problem asks for a function for the integral and it is not f(r,t)*r and wondering why I'm incorrect.

********************************
Well, because you are given f(x,y), and you x≠r and y≠θ; you need to substitute the actual formulas for x and y in terms of r and θ.

mat267 review sessions at math tutoring center

Lecture Notes 2/21/18 through 3/2/18

Lecture Notes 2-21-18 through 3-2-18

11.6#5

I do not understand why this is wrong

******************

 

I don't know what you did since you didn't tell me, and it's hard to tell, but I guess you
computed ∂f/∂y incorrectly and/or then you did the algebra incorrectly when you tried
to normalize the gradient to get the direction vector.