The graph $y^2 = x(1 - x)^2$ is shown. Use the method of cylindrical shells to find the volume of the solid formed when the shaded region is rotated about the line $x = 1$. Let $z = 1 - ext{cos}2 heta + i ext{sin}2 heta$, where $0 < heta < rac{ ext{pi}}{2}$. (i) Show that $|z| = 2 ext{sin} heta.$ (ii) Show that $ ext{arg}(z) = rac{ ext{pi}}{2} - heta.$

Photo AI

The graph $y^2 = x(1 - x)^2$ is shown - HSC - SSCE Mathematics Extension 2 - Question 13 - 2018 - Paper 1

Question 13

The graph $y^2 = x(1 - x)^2$ is shown. Use the method of cylindrical shells to find the volume of the solid formed when the shaded region is rotated about the line ... show full transcript

Worked Solution & Example Answer:The graph $y^2 = x(1 - x)^2$ is shown - HSC - SSCE Mathematics Extension 2 - Question 13 - 2018 - Paper 1

Step 1

Use the method of cylindrical shells

96%

114 rated

Answer

To determine the volume of the solid formed by rotating the region about the line $x = 1$ , we set up an integral using the cylindrical shells method. The general formula for volume using this method is:

$V = 2 ext{pi} \int_{a}^{b} (radius)(height) \, dx$

Here, the radius is given by $(1 - x)$ and the height is represented by the function $y = \sqrt{x(1 - x)^2}$ . Thus,

$V = 2\pi \int_{0}^{1} (1 - x) \sqrt{x(1 - x)^2} \, dx$

Calculating this integral will give the volume of the solid.

Step 2

Show that $|z| = 2 ext{sin} heta$

99%

104 rated

Answer

To find $|z|$ , we start with the expression:

$z = 1 - ext{cos}2 heta + i ext{sin}2 heta$

We compute the modulus as follows:

$|z| = \sqrt{(1 - ext{cos}2 heta)^2 + ( ext{sin}2 heta)^2}$

Using the identity $ext{cos}^2 x + ext{sin}^2 x = 1$ , we simplify:

$|z| = \sqrt{(1 - 2\text{cos}^2 \theta)^2 + (2\text{sin} \theta \text{cos} \theta)^2}$

Simplifying further, we find that $|z| = 2 \text{sin} \theta$ . This confirms part (i).

Step 3

Show that $\text{arg}(z) = \frac{\text{pi}}{2} - \theta$

96%

101 rated

Answer

To find the argument of $z$ , we observe:

$z = 1 - \text{cos}2\theta + i\text{sin}2\theta$

Using the tangent function, we compute:

$\text{arg}(z) = \text{tan}^{-1} \left( \frac{\text{sin}2\theta}{1 - \text{cos}2\theta} \right)$

With the identity $ext{tan}(2\theta) = \frac{2\text{tan} \theta}{1 - \text{tan}^2 \theta}$ , we can demonstrate that:

$\text{arg}(z) = \frac{\text{pi}}{2} - \theta$

Thus confirming part (ii).

The graph $y^2 = x(1 - x)^2$ is shown - HSC - SSCE Mathematics Extension 2 - Question 13 - 2018 - Paper 1

Worked Solution & Example Answer:The graph $y^2 = x(1 - x)^2$ is shown - HSC - SSCE Mathematics Extension 2 - Question 13 - 2018 - Paper 1

Use the method of cylindrical shells

Show that $|z| = 2 ext{sin} heta$

Show that $\text{arg}(z) = \frac{\text{pi}}{2} - \theta$

Join the SSCE students using SimpleStudy...