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2 answers:
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First solve x³=9x
We can ignore x = -3 because it is not in the first quadrant.
<span>So our integral is going to go from x=0 to x=3.
</span>
<span>Now we can use the formula
</span><span>
</span>![\int\limits_{0}^{3}(\pi (9x)^2-\pi(x^3)^2)dx \\\int\limits_{0}^{3}(\pi (9x)^2-\pi(x^3)^2)dx \\\pi\int\limits_{0}^{3}(81x^2-x^6)dx \\2\times \int\limits_{0}^{3}[\pi(9x)^2-\pi(x^3)^2]dx \\\pi\left[81\frac{x^3}{3}-\frac{x^7}{7}\right]_{0}^{3} \\ \\\frac{2916\pi}{7}](https://tex.z-dn.net/?f=%5Cint%5Climits_%7B0%7D%5E%7B3%7D%28%5Cpi%20%289x%29%5E2-%5Cpi%28x%5E3%29%5E2%29dx%0A%5C%5C%5Cint%5Climits_%7B0%7D%5E%7B3%7D%28%5Cpi%20%289x%29%5E2-%5Cpi%28x%5E3%29%5E2%29dx%0A%5C%5C%5Cpi%5Cint%5Climits_%7B0%7D%5E%7B3%7D%2881x%5E2-x%5E6%29dx%0A%5C%5C2%5Ctimes%20%5Cint%5Climits_%7B0%7D%5E%7B3%7D%5B%5Cpi%289x%29%5E2-%5Cpi%28x%5E3%29%5E2%5Ddx%0A%5C%5C%5Cpi%5Cleft%5B81%5Cfrac%7Bx%5E3%7D%7B3%7D-%5Cfrac%7Bx%5E7%7D%7B7%7D%5Cright%5D_%7B0%7D%5E%7B3%7D%20%5C%5C%20%5C%5C%5Cfrac%7B2916%5Cpi%7D%7B7%7D)
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We can ignore x = -3 because it is not in the first quadrant.
<span>So our integral is going to go from x=0 to x=3.
</span>
<span>Now we can use the formula
</span><span>
</span>
</span>