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 \(y=2t\)

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 \(y=2t\)