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Find the area under the curve by using integration:

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My Attempt: Considering a strip of width $dx$ and height $y$ at a distance of $x$ from $Y-$ axis. $$A=\int_{0}^{a} y dx$$ $$=\int_{0}^{a} k(x-a)^{2} dx$$ $$=\int_{0}^{a} \frac {b}{a^2} (x^2-2ax+a^2)dx$$ $$=\frac {b}{a^2} \int_{0}^{a} (x^2-2ax+a^2) dx$$ Thus Area$=\frac {ab}{3}$

Now if we consider a strip of width $dy$ parallel to $X-$ axis $$A=\int_{0}^{b} xdy$$ $$=\int_{0}^{b} \sqrt {\frac {y}{k}}+a dy$$ $$=\frac {1}{\sqrt {k}} (\frac {2}{3} b^{\frac {3}{2}})+ab$$ Thus Area$=\frac {5ab}{3}$

Why am I getting different answers?

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  • $\begingroup$ Because the inverse function of $x\mapsto k(x-a)^2$ is $a-\sqrt{\frac{y}{b}}$ and not what you wrote. Remember that $\sqrt{x^2}\neq |x|$ (and not $x$). $\endgroup$ – Surb Mar 31 '20 at 9:24
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The reason you're getting different answers is you took the wrong sign when you converted the equation of

$$y = k(x-a)^2 \tag{1}\label{eq1A}$$

Note that in the region you're integrating, $x \le a \implies x - a \le 0$. Thus, when you took the square of both sides in your change of variables, you should have used the negative value instead to get

$$x = -\sqrt{\frac{y}{k}} + a \tag{2}\label{eq2A}$$

If you use that, then your integration will work properly. In particular, you would then get

$$\begin{equation}\begin{aligned} A & = \int_{0}^{b} \left(-\sqrt{\frac{y}{k}} + a\right)dy \\ & = -\frac{1}{\sqrt {k}}\left(\frac {2}{3} b^{\frac {3}{2}}\right)+ab \\ & = -\frac{2ab}{3} + ab \\ & = \frac{ab}{3} \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

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  • $\begingroup$ However the correct answer is $\frac {ab}{3}$. Yes? $\endgroup$ – pi-π Mar 31 '20 at 9:56
  • $\begingroup$ @pi-π Yes, the correct answer is $\frac{ab}{3}$. You got that using your first method, and with the correction given in my answer to your second method, you would then also get the same result of $\frac{ab}{3}$. $\endgroup$ – John Omielan Mar 31 '20 at 9:57
  • $\begingroup$ Thank You....... $\endgroup$ – pi-π Mar 31 '20 at 9:59

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