- #1
mathmari
Gold Member
MHB
- 5,049
- 7
Hey! 
I have the following exercise:
Integrate the function $f=x+ \sqrt{y} -z^2$ at the path from $(0,0,0)$ to$(1,1,1)$ that consists of the curves $C_1:r(t)=t \hat{ \imath}+t^2\hat{ \jmath}, 0 \leq t \leq 1$ and $C_2: r(t)=\hat{ \imath}+\hat{ \jmath}+t \hat{k}, 0\leq t \leq1$.My solution is as followed:
$$\int_C{f}ds=\int_{C_1}{f}ds+\int_{C_2}{f}ds$$
$$\int_{C_1}{f}ds=\int_0^1{(t+\sqrt{t^2})\sqrt{1+4t^2}}dt=\int_0^1{2t(1+4t^2)^{\frac{1}{2}}}dt=\frac{1}{6}(1+4t^2)^{\frac{3}{2}}|_0^1=\frac{1}{6}(5 \sqrt{5}-1)$$
$$\int_{C_2}{f}ds=\int_0^1(2-t^2) \cdot 1 dt=[2t-\frac{t^3}{3}]_0^1=\frac{5}{3}$$
So $$\int_C{f}ds=\frac{5 \sqrt{5}-1}{6}+\frac{5}{3}$$Could you tell me if this is correct??
I have the following exercise:
Integrate the function $f=x+ \sqrt{y} -z^2$ at the path from $(0,0,0)$ to$(1,1,1)$ that consists of the curves $C_1:r(t)=t \hat{ \imath}+t^2\hat{ \jmath}, 0 \leq t \leq 1$ and $C_2: r(t)=\hat{ \imath}+\hat{ \jmath}+t \hat{k}, 0\leq t \leq1$.My solution is as followed:
$$\int_C{f}ds=\int_{C_1}{f}ds+\int_{C_2}{f}ds$$
$$\int_{C_1}{f}ds=\int_0^1{(t+\sqrt{t^2})\sqrt{1+4t^2}}dt=\int_0^1{2t(1+4t^2)^{\frac{1}{2}}}dt=\frac{1}{6}(1+4t^2)^{\frac{3}{2}}|_0^1=\frac{1}{6}(5 \sqrt{5}-1)$$
$$\int_{C_2}{f}ds=\int_0^1(2-t^2) \cdot 1 dt=[2t-\frac{t^3}{3}]_0^1=\frac{5}{3}$$
So $$\int_C{f}ds=\frac{5 \sqrt{5}-1}{6}+\frac{5}{3}$$Could you tell me if this is correct??