Maths Problems: Calculus and Integration (1 Viewer)

kudipunjaban · Apr 5, 2009

Hi, I was wondering if anyone could help me with the following:

1) Sketch the curve y=x(x-2)^3, showing any turning points and inflections.
2) Find the maximum possible area if a straight 8m length of fencing is placed across a corner to enclose a triangular space.
3) Find the volume of the solid formed when the curve y= (x+5)^2 is rotated about the y-axis from y=1 to y=4

Thank u

Prashant Sallan · Apr 5, 2009

Hey, this is only Q1

kudipunjaban · Apr 6, 2009

Thank u soooo much.....

Drongoski · Apr 6, 2009

Q3)

Volume formed by rotating the area about the y-axis from y = 1 to 4

$\100dpi y\ =\ (x+5)^{2}\\ \\ \therefore\ \ x\ =\ -5\pm \sqrt{y}\ \ \ i.e.\ \ 2\ branches:\ \ x_{1}\ =\ -5-\sqrt{y}\ \ \ and \ \ x_{2}\ \ =\ \ -5+\sqrt{y}\\ \\ \\ \therefore \ \ Vol\ \ =\ \ V_{1}\ \ -\ \ V_{2}\ \ =\ \ \pi \int_{1}^{4}\ x_{1}^{2}\ dy\ \ -\ \ \pi \int_{1}^{4}\ x_{2}^{2}\ dy\\ \\ \ \ =\ \ \pi \int_{1}^{4}\ ( x_{1}^{2}\ -\ x_{2}^{2})\ dy\ \ =\ \ \pi \int_{1}^{4}\ (-5-\sqrt{y})^{2}\ -\ (-5+\sqrt{y})^{2}\ \ dy\\ \\ \ \ = 20\pi \int_{1}^{4}\ \sqrt{y}\ dy\ \ =\ \ 20\pi \times \frac{2}{3}y^{\frac{3}{2}}]^{4}_{1}\ \ =\ \ \frac{280\pi }{3} \ \ cubic\ units\ \ \\ \\ \\$

Hopefully answer is correct.

jet · Apr 6, 2009

Maths Problems: Calculus and Integration (1 Viewer)

kudipunjaban

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Prashant Sallan

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kudipunjaban

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Drongoski

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jet

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