- #1

Lancelot59

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This particular problem is just confusing me in the setup. I need to find the area that is inside both:

r=sqrt(3)cos(theta) and r=sin(theta)

It makes a petal type shape. I was beating my head around for a while, but I reasoned that since the equation used to find the area cuts out in a straight line. I could just move in either direction following the appropriate functions and get the area by adding the two parts together:

[tex]\frac{1}{2}[\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} (\sqrt{3}cos(\theta))^{2}) d\theta + \int_{0}^{\frac{\pi}{3}} (sin(\theta)^{2} d\theta][/tex]

It makes sense to me, but my final answer was 17pi\4 + 3sqrt(3)/8, however the book states the answer is 5pi/24 - sqrt3/4. Is my setup wrong, or did I just mess up with the integration?

r=sqrt(3)cos(theta) and r=sin(theta)

It makes a petal type shape. I was beating my head around for a while, but I reasoned that since the equation used to find the area cuts out in a straight line. I could just move in either direction following the appropriate functions and get the area by adding the two parts together:

[tex]\frac{1}{2}[\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} (\sqrt{3}cos(\theta))^{2}) d\theta + \int_{0}^{\frac{\pi}{3}} (sin(\theta)^{2} d\theta][/tex]

It makes sense to me, but my final answer was 17pi\4 + 3sqrt(3)/8, however the book states the answer is 5pi/24 - sqrt3/4. Is my setup wrong, or did I just mess up with the integration?

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