# Vector Calc: Can you verify my answer?

Just looking for a yes/no. I worked out the following question:

(A) Calculate the arc length of the curve $$r(t) = (\log t, 2t, t^2)$$ where $$1 \leq t \leq e$$

(B) Let C be the ellipsed form by intersecting the cylinder $$x^2 + y^2 = 1$$ and the plane $$z = 2y + 1$$ and let $$\textbf{f}(x,y,z) = (y,z,x)$$. What is $$\int_C \textbf{f} d\textbf{r}$$.

(C) Let C be the hyperbola formed by intersecting the cone $$x^2 + y^2 = z^2$$ and the plane $$x+y+z=1$$ and let $$\textbf{f}(x,y,z) = \textbf{k}/z^2$$. What is $$\int_C \textbf{f} \times d\textbf{r}$$
I got,

(A) $$e^2$$
(B) $$\pi$$
(C) $$-(1+3\pi/2, 1+3\pi/2, 0)$$

Did I hit the mark?

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HallsofIvy
Homework Helper
I haven't done (C) yet but I don't get anything like you have for either (A) or (B). Show how you got those answers.

I haven't done (C) yet but I don't get anything like you have for either (A) or (B). Show how you got those answers.
Okay. So for (A):

\begin{align*}L &= \int_1^e |r'(t)| dt \\ &= \int_1^e |(1/t, 2, 2t)| dt \\ &= \int_1^e \sqrt{\frac{(2t^2+1)^2}{t^2}} dt = \int_1^e 2t + \frac{1}{t} dt \\ &= e^2 - 1 + 1 = e^2\end{align*}

For (B), the contour is,

$$\gamma(t) = (\cos t, \sin t, 2\cos t + 1)$$

So the relevant integral is,

$$\int_0^{2\pi} (\sin t, 2 \sin t + 1, \cos t) \cdot (-\sin t, \cos t, -2 \cos t) \ dt = \pi$$

Does that look wrong?

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In (B), if you say that y=sin(t), then you should find z=2*y+1=2*sin(t)+1, right?

In (B), if you say that y=sin(t), then you should find z=2*y+1=2*sin(t)+1, right?
Doh! But I think the answer is the same. I've corrected it in the post above.