take $\int_\lambda f \ ds$ (Stieltjes) where $f(x,y) = \sqrt{x^2+y^2}$ and $\lambda(t) = (a\cos t, b\sin t)$ 
Calculate $\int_\lambda f \ ds$, where $f:\mathbb{R}^2\to\mathbb{R}$
  is given by $f(x,y) = \sqrt{x^2+y^2}$ and $\lambda:[0,
 2\pi]\to\mathbb{R^2}$ is the path $\lambda(t) = (a\cos t, b\sin t)$,
  which describes an ellipse.

When $\lambda$ is $C^1$ and $w$ is a continuous $1-form$, we have:
$$\int_\lambda w =\int_a^b w(\lambda(t))\cdot \lambda'(t) dt$$
so:
$$\int_\lambda f\ ds = \int_0^{2\pi} \sqrt{(a\cos t)^2 + (b\sin t)^2}\cdot (-a\sin t, b\cos t) \ dt = \\ \int_0^{2\pi} \sqrt{a^2\cos^2 t + b^2\sin^2 t} \cdot (-a\sin t, b\cos t) \ dt $$
but I can't even take the dot product... I don't hink this is the way to solve.
 A: I studied analysis about twenty years ago, so my answer should be considered not as a rigor proof but as an idea. 
$$\int_\lambda f\ ds = \int_0^{2\pi} f(t)\frac{ds}{dt}dt.$$
Let $\lambda(t)=(\lambda_x(t), \lambda_y(t))$. Then 
$ds=\sqrt{\lambda_x’^2+\lambda_y’^2}dt=\sqrt{a^2\sin^2t+b^2\cos^2 t}dt$,
$$\int_\lambda f\ ds=\int_0^{2\pi} \sqrt{a^2\cos^2 t + b^2\sin^2 t}\cdot 
\sqrt{a^2\sin^2t+b^2\cos^2 t}dt=$$
Now let’s try to find 
$$I=\int \sqrt{a^2\cos^2 t + b^2\sin^2 t}\cdot \sqrt{a^2\sin^2t+b^2\cos^2 t}dt$$
To get rid of trigonometric functions, make a substitution $u=\tan t$. Then $\cos^2 t=\frac{1}{1+\tan^2 t}=\frac{1}{1+u^2}$, $\sin^2 t=\frac{\tan^2 t}{1+\tan^2 t}=\frac{u^2}{1+u^2}$, $t=\operatorname{atan} u$, and $dt=\frac {du}{1+u^2}$. Hence 
$$I=\int \frac{\sqrt{a^2+b^2u^2}\cdot \sqrt{a^2u^2+b^2}}{(1+u^2)^2}du.$$
Then I used for references a fundamental and huge (it has more than two thousand pages in three volumes) book “Differential and Integral Calculus” by Grigorii Fichtenholz. There is written that such integrals are called (pseudo)elliptic and usually cannot be expressed in a finite form by elementary functions even with the extended meaning of this term. Mathcad failed to calculate $I$ too.  
Update. If we interpret the given integral as 
$$\int_\lambda f\ dr =\int_\lambda f\ d\lambda= \int_0^{2\pi} f(t)\frac{d\lambda}{dt}dt.$$
then the calculations become simple. Indeed, let 
$\lambda(t)=(\lambda_x(t), \lambda_y(t))$. Then 
$d\lambda=(\lambda_x’,\lambda_y’)dt=(-a\sin t, b\cos t)dt$,
$$\int_\lambda f\ dr=\int_0^{2\pi} \sqrt{a^2\cos^2 t + b^2\sin^2 t}\cdot 
(-a\sin t, b\cos t)dt.$$
Now let’s try to find 
$$I=\int \sqrt{a^2\cos^2 t + b^2\sin^2 t}\cdot (-a\sin t, b\cos t)dt=(I_1,I_2),$$
where 
$$I_1=\int \sqrt{a^2\cos^2 t + b^2\sin^2 t}(-a\sin t)dt,$$
and $$I_2=\int \sqrt{a^2\cos^2 t + b^2\sin^2 t}(b\cos t)dt.$$
But 
$$I_1=\int \sqrt{a^2\cos^2 t + b^2\sin^2 t}\cdot a\cdot d\cos t=
a\int \sqrt{(a^2-b^2)u^2 + b^2}du,$$
and 
$$I_2=\int \sqrt{a^2\cos^2 t + b^2\sin^2 t}\cdot b\cdot d\sin t=
b\int \sqrt{(b^2-a^2)u^2 + a^2}du$$
At last I recall the standard integrals
$$\int \sqrt{\alpha t^2+\beta}\cdot dt=\frac 12t\sqrt{\alpha t^2+\beta}+\frac{\beta}2\int \frac{dt}{\sqrt{\alpha t^2+\beta}},$$
and 
$$\int \frac{dt}{\sqrt{\alpha t^2+\beta}}=\cases{ 
\frac 1{\sqrt{\alpha}}\ln\left|t+\sqrt{t^2+\frac{\beta}{\alpha}}\right|+C,\mbox{ if }\alpha>0 \\ \frac{1}{\sqrt{|\alpha|}}\operatorname{asin} \left(\sqrt{\frac{|\alpha|}{\beta}}t \right)+C,\mbox{ if }\alpha<0}.$$ 
