- #1
Kaldanis
- 106
- 0
Evaluate the line integral, where C is the given curve.
\int_{c} xy\:ds, when C: x=t^{2}, \ y=2t\ , \ 0\leq t\leq4
To solve this I should use the formula \int^{b}_{a} f(x(t),y(t))\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}dt
This gives me \int^{4}_{0} (t^{2}2t)\sqrt{(2t)^{2}+(2)^{2}}dt
When I solve this integral I get around 860.3. I've had various friends solve it and also checked on my calculator and wolfram, all getting 860.3 each time. I've been told this answer is wrong which leads me to believe I'm using the formula incorrectly. Is my integral above correct? Is it wrong to just plain solve the integral from 0 to 4?
I appreciate any help
\int_{c} xy\:ds, when C: x=t^{2}, \ y=2t\ , \ 0\leq t\leq4
To solve this I should use the formula \int^{b}_{a} f(x(t),y(t))\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}dt
This gives me \int^{4}_{0} (t^{2}2t)\sqrt{(2t)^{2}+(2)^{2}}dt
When I solve this integral I get around 860.3. I've had various friends solve it and also checked on my calculator and wolfram, all getting 860.3 each time. I've been told this answer is wrong which leads me to believe I'm using the formula incorrectly. Is my integral above correct? Is it wrong to just plain solve the integral from 0 to 4?
I appreciate any help
)
