I need some guiding with this Analysis II exercises

Andrei Cosmin

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Jun 8, 2021
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1. Evaluate the following line integral:
∫C ydx - 2xdy , where (C) is the portion of y = x^2 from (2,4) to (0,0)
I applied Green's Theorem but I am not sure if it is the right way to solve it.

2. . Evaluate the triple integral:
∭Q 2y + z dV
Where Q is the solid bounded by z = 6 - x^2 - y^2 and z = 2.
 
For [MATH]1[/MATH]
[MATH]M = y[/MATH]
[MATH]N = -2x[/MATH]
How will you set the integral for those?
 
1. Evaluate the following line integral:
∫C ydx - 2xdy , where (C) is the portion of y = x^2 from (2,4) to (0,0)
I applied Green's Theorem but I am not sure if it is the right way to solve it.[/quote[
Let x= t, y=t2\displaystyle y= t^2 Then dx= dt and dy= 2tdt. As (x, y) goes from (2, 4) to (0, 0), t goes from 2 to 0. The integral is 20t2dt(2t)(2tdt)=20(t24t2)dt=320t2dt=302t2dt\displaystyle \int_2^0 t^2dt- (2t)(2tdt)= \int_2^0 (t^2- 4t^2)dt= -3\int_2^0 t^2dt= 3\int_0^2 t^2dt.

2. . Evaluate the triple integral:
∭Q 2y + z dV
Where Q is the solid bounded by z = 6 - x^2 - y^2 and z = 2.
z=6x2y2\displaystyle z= 6- x^2- y^2 is a paraboloid, opening downward with vertex at (0, 0, 6). It cuts the plane z= 2 where 2=6x2y2\displaystyle 2= 6- x^2- y^2 so where x2+y2=62=4\displaystyle x^2+ y^2= 6- 2= 4. That is the circle with center at (0, 0, 2) and radius 2.

I would use cylindrical coordinates. Since the "base" at z= 2 is the full circle, θ\displaystyle \theta goes from 0 to 2π\displaystyle 2\pi. Since the radius of that circle is 2, r goes from 0 to 2. And, of course, z goes from 2 to 6x2y2=6r2\displaystyle 6- x^2- y^2= 6- r^2.

As a triple integral that is 02π0206r2rdzdrdθ\displaystyle \int_0^{2\pi}\int_0^2\int_0^{6- r^2} r dzdrd\theta.
 
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