Question

Calculus 3 help please.​

Answer 1

I assume each path
`C` is oriented positively/counterclockwise.

(a) Parameterize
`C` by

`\begin{cases} x(t) = 4\cos(t) \\ y(t) = 4\sin(t)\end{cases} \implies \begin{cases} x'(t) = -4\sin(t) \\ y'(t) = 4\cos(t) \end{cases}`

with
`-\frac\pi2\le t\le\frac\pi2`. Then the line element is

`ds = √(x'(t)^2 + y'(t)^2) \, dt = √(16(\sin^2(t)+\cos^2(t))) \, dt = 4\,dt`

and the integral reduces to

`\displaystyle \int_C xy^4 \, ds = \int_(-\pi/2)^(\pi/2) (4\cos(t)) (4\sin(t))^4 (4\,dt) = 4^6 \int_(-\pi/2)^(\pi/2) \cos(t) \sin^4(t) \, dt`

The integrand is symmetric about
`t=0`, so

`\displaystyle 4^6 \int_(-\pi/2)^(\pi/2) \cos(t) \sin^4(t) \, dt = 2^(13) \int_0^(\pi/2) \cos(t) \sin^4(t) \,dt`

Substitute
`u=\sin(t)` and
`du=\cos(t)\,dt`. Then we get

`\displaystyle 2^(13) \int_0^(\pi/2) \cos(t) \sin^4(t) \, dt = 2^(13) \int_0^1 u^4 \, du = \frac{2^(13)}5 (1^5 - 0^5) = \boxed{\frac{8192}5}`

(b) Parameterize
`C` by

`\begin{cases} x(t) = 2(1-t) + 5t = 3t - 2 \\ y(t) = 0(1-t) + 4t = 4t \end{cases} \implies \begin{cases} x'(t) = 3 \\ y'(t) = 4 \end{cases}`

with
`0\le t\le1`. Then

`ds = √(3^2+4^2) \, dt = 5\,dt`

and

`\displaystyle \int_C x e^y \, ds = \int_0^1 (3t-2) e^(4t) (5\,dt) = 5 \int_0^1 (3t - 2) e^(4t) \, dt`

Integrate by parts with

`u = 3t-2 \implies du = 3\,dt \\\\ dv = e^(4t) \, dt \implies v = \frac14 e^(4t)`

`\displaystyle \int u\,dv = uv - \int v\,du`

`\implies \displaystyle 5 \int_0^1 (3t-2) e^(4t) \,dt = \frac54 (3t-2) e^(4t) \bigg|_(t=0)^(t=1) - \frac{15}4 \int_0^1 e^(4t) \,dt \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - (15)/(16) e^(4t) \bigg|_(t=0)^(t=1) \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - (15)/(16) (e^4 - 1) = \boxed{(5e^4 + 55)/(16)}`

(c) Parameterize
`C` by

`\begin{cases} x(t) = 3(1-t)+t = -2t+3 \\ y(t) = (1-t)+2t = t+1 \\ z(t) = 2(1-t)+5t = 3t+2 \end{cases} \implies \begin{cases} x'(t) = -2 \\ y'(t) = 1 \\ z'(t) = 3 \end{cases}`

with
`0\le t\le1`. Then

`ds = √((-2)^2 + 1^2 + 3^2) \, dt = √(14) \, dt`

and

`\displaystyle \int_C y^2 z \, ds = \int_0^1 (t+1)^2 (3t+2) \left(√(14)\,ds\right) \\\\ ~~~~~~~~ = √(14) \int_0^1 \left(3t^3 + 8t^2 + 7t + 2\right) \, dt \\\\ ~~~~~~~~ = √(14) \left(\frac34 t^4 + \frac83 t^3 + \frac72 t^2 + 2t\right) \bigg|_(t=0)^(t=1) \\\\ ~~~~~~~~ = √(14) \left(\frac34 + \frac83 + \frac72 + 2\right) = \boxed{(107√(14))/(12)}`