Question

I need help with my work

Answer 1

The area of the interior above the polar axis is -0.858 square units

The area bounded by a polar curve

The area bounded by a polar curve between θ = θ₁ and θ = θ₂ is given by

$A = \int\limits^{\theta_{2} }_{\theta_{1} } {\frac{1}{2}r^{2} } \, d\theta$

Now, since we have the curve r = 1 - sinθ and we want to find the area of the interior above the polar axis, we integrate from θ = 0 to θ = π, since this is the region above the polar axis.

So, $A = \int\limits^{\theta_{2} }_{\theta_{1} } {\frac{1}{2}r^{2} } \, d\theta\\= \int\limits^{\pi}_{0} {\frac{1}{2}(1 - sin\theta)^{2} } \, d\theta\\= \int\limits^{\pi}_{0} {\frac{1}{2}[1 - 2sin\theta + (sin\theta)^{2}] } \, d\theta\\= \int\limits^{\pi}_{0} {\frac{1}{2}\, d\theta - \int\limits^{\pi}_{0} 2sin\theta \, d\theta+ \int\limits^{\pi}_{0} (sin\theta)^{2} } \, d\theta$

$A = \int\limits^{\pi}_{0} {\frac{1}{2}\, d\theta - 2\int\limits^{\pi}_{0} sin\theta \, d\theta+ \int\limits^{\pi}_{0} \frac{(1 - cos2\theta)}{2} } \, d\theta\\= \int\limits^{\pi}_{0} {\frac{1}{2}\, d\theta - 2\int\limits^{\pi}_{0} sin\theta \, d\theta+ \int\limits^{\pi}_{0} \frac{1}{2} } \, d\theta -\int\limits^{\pi}_{0} \frac{cos2\theta}{2} } \, d\theta$

$A = \int\limits^{\pi}_{0} \, d\theta - 2\int\limits^{\pi}_{0} sin\theta \, d\theta -\int\limits^{\pi}_{0} \frac{cos2\theta}{2} } \, d\theta\\= [\theta]_{0}^{\pi} - 2[-cos\theta]^{\pi} _{0} - [\frac{sin2\theta}{4}]_{0}^{\pi} \\= [\theta]_{0}^{\pi} + 2[cos\theta]^{\pi} _{0} - [\frac{sin2\theta}{4}]_{0}^{\pi}\\= [\pi - 0] + 2[cos\pi - cos0] - \frac{ [sin2\pi - sin0]}{4}\\= \pi + 2[-1 - 1] - \frac{ [0 - 0]}{4}\\= \pi + 2[-2] - \frac{ [0]}{4}\\= \pi - 4 - 0\\= \pi - 4\\= 3.142 - 4\\= -0.858$

So, the area of the interior above the polar axis is -0.858 square units

Learn more about area bounded by polar curve here:

brainly.com/question/27624501

#SPJ1