Question

Evaluate the line integral \int_c xyds Where c is the given curve c:x=t^2,y=2t,0\leq t\leq5

Integrals
ANSWERED
asked 2021-06-07
Evaluate the line integral
\(\int_c xyds\)
Where c is the given curve
\(c:x=t^2,y=2t,0\leq t\leq5\)

Expert Answers (1)

2021-06-08
Explanation
Line integrals can be used to eavaluate integrals of two or three dimensioal curves
The curve c is expressed by the following parametric equations given by
\(x=t^2\)
and
\(y=2t\)
Thus, to evaluate the following integral given by
\(I=\int_c xyds\)
we use the following line integral formula given by
\(\int_c f(x,y)ds=\int_a^b f(x(t),y(t))|r'|dt\)
\(=\int_a^b f(x(t),y(t))|\frac{dr}{dt}|dt\)
\(=\int_a^b f(x(t),y(t))\sqrt{(\frac{dx}{dt})2+(\frac{dy}{dt})^2}dt\)
Since \(x(t)=t^2\) and \(y(t)=2t\)
We want to evaluate \(\int_c xyds\) as follows
Firstly, we must find the integrand \(f(x(t),y(t))\) as follows
\(f(x(t),y(t))=xy\)
\(=t^2(2t)\)
\(=2t^3\)
Also, the derivatives of x(f) and y(t) must be
\(\frac{d}{dt}x(t)=\frac{d}{dt}(t^2)=2t\)
and
\(\frac{d}{dt}y(t)=\frac{d}{dt}(2t)=2\)
Since \(0\leq t\leq5\), so that the upper and lower limits of the integral must be
\(a=0\)
and
\(b=5\)
Thus, using the line integral formula, we get
\(\int_c xyds=\int_a^b f(x(t))\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt\)
\(=\int_0^5(2t^3)\sqrt{(2t)^2+(2)^2}dt\)
\(=2\int_0^5t^3\sqrt{4t^2+4}dt=2\int_0^5 t^3\sqrt{4(t^2+1)}dt\)
\(=2\int_0^5 2t^3\sqrt{(t^2+1)}dt\)
\(=4\int_0^5 t^3\sqrt{(t^2+1)}dt\)
To evaluate the following integral \(\int_0^5 t^3\sqrt{(t^2+1)}dt\), we use the method of substitution as follows
Let \(u=t^2+1\), so that
Differentiate both sides of the following equation \(u=t^2+1\) implies that
\(du=2tdt\)
and thus
\(tdt=\frac{du}{2}\)
Also, we have
\(t^2=u-1\)
Also, we find that the upper and lower limits of the integral must be
\(t=0\Rightarrow u=0^2+1=1\)
and
\(t=5\Rightarrow u=5^2+1=26\)
Thus, the line integral becomes
\(\int_c xyds=4\int_{t=0}^5 t^2\sqrt{t^2+1}tdt\)
\(=4\int_{u=1}^{26}(u-1)\sqrt{u}(\frac{du}{2})\)
\(=\frac{4}{2}\int_{u=1}^{26}(u-1)\sqrt{u}du\)
\(=2\int_{u=1}^{26}(u^{\frac{3}{2}}-u^\frac{1}{2})du\)
Thus, evaluating the following line integral implies that
\(\int_c xyds=2[\frac{u^{\frac{5}{2}}}{\frac{5}{2}}-\frac{u^{\frac{3}{2}}}{\frac{3}{2}}]_{u=1}^{26}\)
\(=2[\frac{2u^{\frac{5}{2}}}{5}-\frac{2u^{\frac{3}{2}}}{3}]\)
\(=2[(\frac{2(26)^\frac{5}{2}}{5}-\frac{2(26)^\frac{3}{2}}{3})-(\frac{2(1)^\frac{5}{2}}{5}-\frac{2(1)^\frac{3}{2}}{3})]\)
\(=\frac{8}{15}(949\sqrt{26}+1)\)
Result:
We have shown that the following line integral implies that
\(\int_c xyds=\frac{8}{15}(949\sqrt{26}+1)\)
for the curve c expressed by the following parametric equations given by
\(x=t^2\)
and \(y=2t\)
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