$$\begin{align} &\text{Power collected on earth} \\ &= \int I\ dA \\ &= \int_0^{R_e} \frac{P_0}{4\pi[(R_0 + R_e - \sqrt {R_e^2 - R^2})^2 + R^2]} 2\pi R\ dR \\ &= \frac{P_0}{2} \int_0^{R_e} \frac{R}{ 2R^2 - 2(R_0+R_e)\sqrt{R_e^2 - R^2} + R_e^2 + (R_0 + R_e)^2}\ dR \end{align}$$

The only variable here is R. But would like the answer to contain all the other constants, so I may proceed with further manipulations later. Thank you, and hope to get some help from the mathematicians present here.

For those who are interested in the physics derivation, here it is, step by step (upon request by @hardmath):

Consider the Earth located at a distance of Ro from the sun. Consider the sun as a point source of power, emitting power of $P_0$.

Intensity at every point on the surface of the earth is not the same, as each point on the earth is located at a different distance from the sun. Only concentric rings are located the same distance from the sun, and thus will capture the same intensity.

Intensity expression:

Let the radius of a concentric ring be $R$. Let the depth of the concentric ring from the surface of earth be $x$.

Intensity, $$I = \frac{P_0}{4\pi[(R_0 + x)^2 + R^2)]} \tag{1}$$

Finding an expression for $x$ in terms of $R$:

$$R_e^2 = R^2 + (R_e - x)^2,$$ where $R_e$ is the radius of the Earth

$$x = R_e - \sqrt {R_e^2 - R^2} \tag{2}$$

Subst (2) into (1): $$I = \frac{P_0}{4\pi[(R_0 + R_e - \sqrt {R_e^2 - R^2})^2 + R^2]} \tag{3}$$

Area expression:

Next, I find an expression for area of an infinitesimally thin concentric ring, $dA$. I can slice the concentric ring once, to make it into a rectangle.

Thus, area of concentric ring, $$dA = 2\pi R\ dR \tag{4}$$

Hence, the power collected can be calculated using the expression above.

  • $\begingroup$ I tried calculating using online calculators, but none of them were powerful enough to do so. Require some human help. $\endgroup$
    – Anonymous
    Oct 24 '21 at 10:33
  • $\begingroup$ Can you please format this properly? $\endgroup$
    – Anixx
    Oct 24 '21 at 12:10
  • $\begingroup$ Alright, done it. Please help to answer. $\endgroup$
    – Anonymous
    Oct 24 '21 at 12:11
  • $\begingroup$ The body of your Question should provide a self-contained problem statement, but here the only indication that you wanted help with performing the indicated integration is in the title, "How do I integrate this expression?". You barely hint at your approach and interest in the problem in the body ("would like the answer to contain all the other constants"), and it would help Readers to know more of this context. $\endgroup$
    – hardmath
    Oct 26 '21 at 0:59
  • $\begingroup$ Thanks for the input. Will edit. $\endgroup$
    – Anonymous
    Oct 26 '21 at 1:44

For the sake of typing and also to make the solution look neat, let $$a=2(R_0+R_e)$$ $$b=R_e$$ $$c=R_e^2+(R_0+R_e)^2$$

So, the required integral becomes $$\int_{0}^{R_e}\frac{R\space dR}{2R^2-a\sqrt{b^2-R^2}+c}$$

Substitute $b^2-R^2=t^2$.

$\Rightarrow R\space dR=-t\space dt$

At $R=0$, we have $t=b=R_e$

At $R=R_e$, we have $t=b^2-R_e^2=0$. \begin{align} \Rightarrow\int_{0}^{R_e}\frac{R\space dR}{2R^2-a\sqrt{b^2-r^2}+c} &= \int_{R_e}^{0}\frac{-t\space dt}{2(b^2-t^2)-at+c}\\ &=\int_{R_e}^{0}\frac{t\space dt}{2t^2+at-(2b^2+c)}\\ \end{align}

This is a pretty standard integral of the form $\int\frac{Linear}{Quadratic}$ which we solve by expressing $\text{Numerator}=l(\text{Derivative of Denominator})+m$ where $l$ and $m$ are constants. This process reduces the original integral into two simpler integrals which can be solved easily.

  • $\begingroup$ Not a mathematician, just a student. $\endgroup$
    – RiverX15
    Oct 24 '21 at 13:56
  • $\begingroup$ This is fantastic. How were you able to decide which substitution to use? $\endgroup$
    – Anonymous
    Oct 24 '21 at 14:03
  • $\begingroup$ @Anonymous The original integral has the term $\sqrt{b^2-R^2}$ and another term of $R^2$. Also, the numerator has the term $R\space dR$ which is just $\frac 12 d(R^2)$. The second substitution also has a similar explanation, though it's not very intuitive at first glance. $\endgroup$
    – RiverX15
    Oct 24 '21 at 14:18

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