Integrate the function $w=x+y^2$

Question

I have the following exercise:

We want to integrate the function $w=x+y^2$ and we have a path that begins from $A(0,0)$ and reaches at $B(1,1)$.

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Could you give me some hint what I am supposed to do?

Do I have to use the following formula? $$\int_C w ds= \int_a^b w(x(t), y(t), z(t)) \sqrt{x'^2(t)+y'^2(t)+z'^2(t)}dt$$

But how??

Answer 1

Yes! That's the formula to use, which integrates a function along a curve.

You're supposed to think up a formula for a (presumably straight) path $(x(t), y(t), z(t))$ that starts at A and ends at B. When you have it, you're supposed to substitute it in the formula and evaluate.

Answer 2

Without knowing the path you're using, it is impossible to get very specific.

The idea here is that your path can be parameterized by two continuous functions $x,y:[a,b]\to\Bbb R$ such that:

• $x,y$ differentiable on $(a,b),$
• $x(a),y(a)=0$ and $x(b),y(b)=1,$
• every point on the path $C$ is of the form $\bigl(x(t),y(t)\bigr)$ for some $a\le t\le b,$ and
• every point of the form $\bigl(x(t),y(t)\bigr)$ for some $a\le t\le b$ lies on the path $C.$

Once you've found such a parameterization, you will use the formula $$\int_Cw\,ds=\int_a^bw\bigl(x(t),y(t)\bigr)\cdot\sqrt{x'(t)^2+y'(t)^2}\,dt.\tag{$\star$}$$

In general, though, things will depend on the path (though not the parameterization).

For example, let's suppose that you wish to move from $(0,0)$ to $(1,1)$ along the curve $y=x.$ One way we can parameterize this is to let $x(t)=t,y(t)=t$ for $t\in[0,1].$ Then by $(\star)$ we have $$\int_Cw\,ds=\int_0^1(t+t^2)\sqrt{1^2+1^2}\,dt=\sqrt2\int_0^1(t^2+t)\,dt=\frac{5\sqrt2}6.$$ Alternately, we could (for example) let $x(t)=t^2,y(t)=t^2$ for $t\in[0,1].$ Then by $(\star)$ we have $$\int_Cw\,ds=\int_0^1(t^2+(t^2)^2)\sqrt{(2t)^2+(2t)^2}\,dt=\int_0^1(t^4+t^2)\sqrt{8t^2}\,dt=2\sqrt2\int_0^1(t^5+t^3)=\frac{5\sqrt2}6.$$ As I mentioned before, it doesn't matter how you parameterize a path--you should get the same value.

On the other hand, suppose we wish to move from $(0,0)$ to $(1,1)$ along the path $y=\sqrt x$--equivalently, $x=y^2.$ Letting $x(t)=t^2$ and $y(t)=t$ for $t\in[0,1],$ we have by $(\star)$ that $$\int_Cw\,ds=\int_0^1(t^2+t^2)\sqrt{(2t)^2+1^2}\,dt=2\int_0^1t^2\sqrt{4t^2+1}\,dt$$ which can be shown to have a value of about $1.21267,$ which is different from the value found along the other path.