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The following problem is from the book, Calculus and Analytical Geometer by Thomas and Finney.
Problem:
Find the area of the region bounded by the given curves. $$ y^2 = 9x, y = \frac{3x^2}{8} $$
Answer:
\begin{align*} 9x &= \frac{3x^2}{8} \\ x &= 0 \,\, \text{ is one solution } \\ \frac{3x^2}{8} &= 9 \\ 3x^2 &= 72 \\ x &= \sqrt{24} = \sqrt{6(4)} \\ x &= 2 \sqrt{6} \\ \end{align*} Let $A$ be the area we seek. \begin{align*} A &= \int_{0}^{2 \sqrt{6}} 3\sqrt{x} - \frac{3x^2}{8} \, dx \\ A &= 3 \left( \frac{2}{3} \right) x^{\frac{3}{2}} - \frac{x^3}{8} \Big|_{0}^{2 \sqrt{6}} \\ A &= 2 x^{\frac{3}{2}} - \frac{x^3}{8} \Big|_{0}^{2 \sqrt{6}} \\ A &= 2 \left( 2 \sqrt{6} \right) ^{ \frac{3}{2} } - \frac{8(6\sqrt{6})}{8} \\ A &= 4 \sqrt{2}(6^\frac{3}{4}) - 6\sqrt{6} \\ \end{align*} The book's answer is $8$. Where did I go wrong?

Based upon the group's comments, I have updated my answer. Here it is: \begin{align*} 9x &= \frac{3x^2}{8} \\ x &= 0 \,\, \text{ is one solution } \\ \frac{3x}{8} &= 9 \\ 3x &= 72 \\ x &= 24 \\ \end{align*} Let $A$ be the area we seek. \begin{align*} A &= \int_{0}^{24} 3\sqrt{x} - \frac{3x^2}{8} \, dx \\ A &= 3 \left( \frac{2}{3} \right) x^{\frac{3}{2}} - \frac{x^3}{8} \Big|_{0}^{24} \\ A &= 2 x^{\frac{3}{2}} - \frac{x^3}{8} \Big|_{0}^{24} \\ A &= 2(24) \sqrt{24} - \frac{24(24)(24)}{8} = 48 \sqrt{24} - 3(24)(24) \\ \end{align*}

However, I am still getting the wrong answer. I believe that I have the wrong curve on top and therefore, I am getting the wrong answer. That is not my only error.

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    $\begingroup$ In the third line of your answer, you forgot to cancel an $x$ term. $\endgroup$ – JG123 Aug 16 at 22:42
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There is an error in the calculation of the upper bound, where a square operation is overlooked. The correct equation for the bounds is,

$$9x = \left(\frac{3}{8}x^2\right)^2$$

which yields $x_1=0$ and $x_2=4$. As a result,

$$\int_0^4 \left( 3\sqrt{x} - \frac{3}{8}x^2 \right) = 8 $$

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You have the right idea in that you have to first find the points of intersection of the two curves, but you are forgetting that one curve was given as $y^2$ and the other was given as $y$. Try rewriting the first curve as $y=\pm3\sqrt{x}$, $x\geq 0$, and try again. Your overall method is correct!

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Note that you have $y^2=9x $ and $y= \frac{3x^2}{8}$ thus you need to have $$9x = (\frac{3x^2}{8})^2 $$ to start with.

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  • $\begingroup$ Sir you make error 3x/8 be (3x/8)^2 $\endgroup$ – yuvraj singh Aug 17 at 6:06

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