What is the solution of the equation (x – 5)^2 + 3(x – 5) + 9 = 0? Use u substitution and the quadratic formula to solve.
1 answer:
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We draw region ABC. Lines that connect y = 0 and y = x³ are vertical so:
(i) prependicular to the axis x - disc method;
(ii) parallel to the axis y - shell method;
(iii) parallel to the line x = 18 - shell method.
Limits of integration for x are easy x₁ = 0 and x₂ = 9.
Now, we have all information, so we could calculate volume.
(i)
![V=\pi\cdot\int\limits_0^9(x^3)^2\, dx=\pi\cdot\int\limits_0^9x^6\, dx=\pi\cdot\left[\dfrac{x^7}{7}\right]_0^9=\pi\cdot\left(\dfrac{9^7}{7}-\dfrac{0^7}{7}\right)=\dfrac{9^7}{7}\pi=\\\\\\=\boxed{\dfrac{4782969}{7}\pi}](https://tex.z-dn.net/?f=V%3D%5Cpi%5Ccdot%5Cint%5Climits_0%5E9%28x%5E3%29%5E2%5C%2C%20dx%3D%5Cpi%5Ccdot%5Cint%5Climits_0%5E9x%5E6%5C%2C%20dx%3D%5Cpi%5Ccdot%5Cleft%5B%5Cdfrac%7Bx%5E7%7D%7B7%7D%5Cright%5D_0%5E9%3D%5Cpi%5Ccdot%5Cleft%28%5Cdfrac%7B9%5E7%7D%7B7%7D-%5Cdfrac%7B0%5E7%7D%7B7%7D%5Cright%29%3D%5Cdfrac%7B9%5E7%7D%7B7%7D%5Cpi%3D%5C%5C%5C%5C%5C%5C%3D%5Cboxed%7B%5Cdfrac%7B4782969%7D%7B7%7D%5Cpi%7D)
Answer B. or D.
(ii)
![V=2\pi\cdot\int\limits_0^{9}(x\cdot x^3)\, dx=2\pi\cdot\int\limits_0^{9}x^4\, dx= 2\pi\cdot\left[\dfrac{x^5}{5}\right]_0^9=2\pi\cdot\left(\dfrac{9^5}{5}-\dfrac{0^5}{5}\right)=\\\\\\=2\pi\cdot\dfrac{9^5}{5}=\boxed{\dfrac{118098}{5}\pi}](https://tex.z-dn.net/?f=V%3D2%5Cpi%5Ccdot%5Cint%5Climits_0%5E%7B9%7D%28x%5Ccdot%20x%5E3%29%5C%2C%20dx%3D2%5Cpi%5Ccdot%5Cint%5Climits_0%5E%7B9%7Dx%5E4%5C%2C%20dx%3D%0A2%5Cpi%5Ccdot%5Cleft%5B%5Cdfrac%7Bx%5E5%7D%7B5%7D%5Cright%5D_0%5E9%3D2%5Cpi%5Ccdot%5Cleft%28%5Cdfrac%7B9%5E5%7D%7B5%7D-%5Cdfrac%7B0%5E5%7D%7B5%7D%5Cright%29%3D%5C%5C%5C%5C%5C%5C%3D2%5Cpi%5Ccdot%5Cdfrac%7B9%5E5%7D%7B5%7D%3D%5Cboxed%7B%5Cdfrac%7B118098%7D%7B5%7D%5Cpi%7D)
So we know that the correct answer is D.
(iii)
Line x = h
![V=2\pi\cdot\int\limits_0^9\big((18-x)\cdot x^3\big)\, dx=2\pi\cdot\int\limits_0^9(18x^3-x^4)\, dx=\\\\\\=2\pi\cdot\left(\int\limits_0^918x^3\, dx-\int\limits_0^9x^4\, dx\right)=2\pi\cdot\left(18\int\limits_0^9x^3\, dx-\int\limits_0^9x^4\, dx\right)=\\\\\\=2\pi\cdot\left(18\left[\dfrac{x^4}{4}\right]_0^9-\left[\dfrac{x^5}{5}\right]_0^9\right)=2\pi\cdot\Biggl(18\biggl(\dfrac{9^4}{4}-\dfrac{0^4}{4}\biggr)-\biggl(\dfrac{9^5}{5}-\dfrac{0^5}{5}\biggr)\Biggr)=\\\\\\](https://tex.z-dn.net/?f=V%3D2%5Cpi%5Ccdot%5Cint%5Climits_0%5E9%5Cbig%28%2818-x%29%5Ccdot%20x%5E3%5Cbig%29%5C%2C%20dx%3D2%5Cpi%5Ccdot%5Cint%5Climits_0%5E9%2818x%5E3-x%5E4%29%5C%2C%20dx%3D%5C%5C%5C%5C%5C%5C%3D2%5Cpi%5Ccdot%5Cleft%28%5Cint%5Climits_0%5E918x%5E3%5C%2C%20dx-%5Cint%5Climits_0%5E9x%5E4%5C%2C%20dx%5Cright%29%3D2%5Cpi%5Ccdot%5Cleft%2818%5Cint%5Climits_0%5E9x%5E3%5C%2C%20dx-%5Cint%5Climits_0%5E9x%5E4%5C%2C%20dx%5Cright%29%3D%5C%5C%5C%5C%5C%5C%3D2%5Cpi%5Ccdot%5Cleft%2818%5Cleft%5B%5Cdfrac%7Bx%5E4%7D%7B4%7D%5Cright%5D_0%5E9-%5Cleft%5B%5Cdfrac%7Bx%5E5%7D%7B5%7D%5Cright%5D_0%5E9%5Cright%29%3D2%5Cpi%5Ccdot%5CBiggl%2818%5Cbiggl%28%5Cdfrac%7B9%5E4%7D%7B4%7D-%5Cdfrac%7B0%5E4%7D%7B4%7D%5Cbiggr%29-%5Cbiggl%28%5Cdfrac%7B9%5E5%7D%7B5%7D-%5Cdfrac%7B0%5E5%7D%7B5%7D%5Cbiggr%29%5CBiggr%29%3D%5C%5C%5C%5C%5C%5C)
Answer D. just as before.
(i) prependicular to the axis x - disc method;
(ii) parallel to the axis y - shell method;
(iii) parallel to the line x = 18 - shell method.
Limits of integration for x are easy x₁ = 0 and x₂ = 9.
Now, we have all information, so we could calculate volume.
(i)
Answer B. or D.
(ii)
So we know that the correct answer is D.
(iii)
Line x = h
Answer D. just as before.