Question

# Evaluate the line integral, where C is the given curve. \int_{C}xy\ ds C:x=t^{2}, y=2t, 0\leq t\leq4

Integrals
Evaluate the line integral, where C is the given curve.
$$\int_{C}xy\ ds$$
$$C:x=t^{2},$$
$$y=2t,$$
$$0\leq t\leq4$$

2021-05-10

Step 1
Consider
$$x=t^{2}$$
$$\frac{dx}{dt}=2t$$
$$y=2t$$
$$\frac{dy}{dt}=2$$
Consider:
$$ds=\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}\cdot dt$$
$$=\sqrt{(2t)^{2}+(2)^{2}}\cdot dt$$
$$=\sqrt{4t^{2}+4}\cdot dt$$
$$=\sqrt{4(t^{2}+1)}\cdot dt$$
$$ds=2\sqrt{(t^{2}+1)}\cdot dt$$
Hence the line integral is,
$$\int_{C}xy\ ds=\int_{t=0}^{4}(t)^{2}(2t)(2\sqrt{t^{2}+1}\ dt)$$
$$=4\int_{t=0}^{4}t^{3}\sqrt{t^{2}+1}\ dt$$
1) $$=4\int_{t=0}^{4}t^{2}\sqrt{t^{2}+1}t\cdot dt$$
Step 2
Let
$$t^{2}+1=u^{2}$$
$$2t\ dt=2u\ du$$
$$t\ dt=u\ du$$
For
$$t=0$$ then
$$t^{2}+1=u^{2}$$
$$u^{2}=1$$
$$u=1$$
For
$$t=4$$ then
$$t^{2}+1=u^{2}$$
$$u^{2}=4^{2}+1$$
$$u^{2}=17$$
$$u=\sqrt{17}$$
Substitute these values in (1) then
$$\int_{C}xy\ ds=4\int_{t=0}^{4}t^{2}\sqrt{t^{2}+1}t\cdot dt$$
$$=4\int_{u=1}^{\sqrt{17}}(\sqrt{u^{2}})\times u\ du$$ $$=4\int_{u=1}^{\sqrt{17}}(u^{2}-1)u\times u\ du$$
$$=4\int_{u=1}^{\sqrt{17}}(u^{2}-1)u^{2}\ du$$
$$=4\int_{u=1}^{\sqrt{17}}u^{4}-u^{2}\ du$$
$$=4\left[\frac{u^{5}}{5}-\frac{u^{3}}{3}\right]_{1}^{\sqrt{17}}$$
$$=4\left[\left(\frac{(\sqrt{17})^{5}}{5}-\frac{(\sqrt{17})^{3}}{3}\right)-\left(\frac{1}{5}-\frac{1}{3}\right)\right]$$
$$=4\left(\frac{782\sqrt{17}}{15}-\left(-\frac{2}{15}\right)\right)$$
$$\int_{C}xy\ ds=860.33$$