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Help calculating area inside circle and outside cardioid
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I've calculated the area inside the circle $r=3acos(theta)$ and outside the cardioid $r=a(1+cos(theta))$ and I got two answers: $a^2pi + a^2frac{sqrt{3}}{2}$ and the answer: $a^2 pi$. Can someone please help? Because I'm not quite sure what the correct answer is.
calculus polar-coordinates
edited Apr 18 '16 at 18:11
Jennifer
8,41721737
asked Apr 18 '16 at 17:40
user332320user332320
2216
$endgroup$
add a comment |
$begingroup$
I've calculated the area inside the circle $r=3acos(theta)$ and outside the cardioid $r=a(1+cos(theta))$ and I got two answers: $a^2pi + a^2frac{sqrt{3}}{2}$ and the answer: $a^2 pi$. Can someone please help? Because I'm not quite sure what the correct answer is.
calculus polar-coordinates
edited Apr 18 '16 at 18:11
Jennifer
8,41721737
asked Apr 18 '16 at 17:40
user332320user332320
2216
$endgroup$
add a comment |
$begingroup$
I've calculated the area inside the circle $r=3acos(theta)$ and outside the cardioid $r=a(1+cos(theta))$ and I got two answers: $a^2pi + a^2frac{sqrt{3}}{2}$ and the answer: $a^2 pi$. Can someone please help? Because I'm not quite sure what the correct answer is.
calculus polar-coordinates
edited Apr 18 '16 at 18:11
Jennifer
8,41721737
asked Apr 18 '16 at 17:40
user332320user332320
2216
$endgroup$
I've calculated the area inside the circle $r=3acos(theta)$ and outside the cardioid $r=a(1+cos(theta))$ and I got two answers: $a^2pi + a^2frac{sqrt{3}}{2}$ and the answer: $a^2 pi$. Can someone please help? Because I'm not quite sure what the correct answer is.
calculus polar-coordinates
calculus polar-coordinates
edited Apr 18 '16 at 18:11
Jennifer
8,41721737
asked Apr 18 '16 at 17:40
user332320user332320
2216
edited Apr 18 '16 at 18:11
Jennifer
8,41721737
asked Apr 18 '16 at 17:40
user332320user332320
2216
edited Apr 18 '16 at 18:11
Jennifer
8,41721737
edited Apr 18 '16 at 18:11
Jennifer
8,41721737
edited Apr 18 '16 at 18:11
Jennifer
8,41721737
8,41721737
asked Apr 18 '16 at 17:40
user332320user332320
2216
asked Apr 18 '16 at 17:40
user332320user332320
2216
asked Apr 18 '16 at 17:40
user332320user332320
2216
2216
add a comment |
add a comment |
1 Answer
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$begingroup$
If $$r_1(theta) = 3a cos theta$$ is the equation of the circle, and $$r_2theta) = a(1+cos theta)$$ is the equation for the cardioid, then these curves intersect when $r_1(theta) = r_2(theta)$, or $3a cos theta = a(1+cos theta)$, or $$2 cos theta = 1.$$ This gives us the intersection points $$theta = pm frac{pi}{3}.$$ There is another point of intersection at the origin, because the circle is traversed twice for a single traversal of the cardioid. But for our purposes, we can integrate on $theta in [-pi/3, pi/3]$. The area enclosed between $r_1$ and $r_2$ is then given by $$A = int_{theta = -pi/3}^{pi/3} frac{1}{2}left(r_1^2(theta) - r_2^2(theta)right) , dtheta.$$ Note this is not the same as $$int_{theta = -pi/3}^{pi/2} frac{1}{2}left(r_1(theta) - r_2(theta)right)^2 , dtheta.$$ The second integral is incorrect for the area enclosed.
answered Apr 18 '16 at 18:01
heropupheropup
62.9k66099
$endgroup$
$begingroup$
I have used your answer and was left with a^2 pi + a^2 (sqrt(3)/2). Is that correct?
$endgroup$
– user332320
Apr 18 '16 at 18:27
$begingroup$
@user332320 No, it is not correct. $$r_1^2(theta) - r_2^2(theta) = 9a^2 cos^2 theta - a^2 (1+costheta)^2 = a^2 (8 cos^2 theta - 2 cos theta - 1).$$ The antiderivative is $$a^2 (2 sin 2theta - 2 sin theta + 3theta).$$ Evaluating that at $theta = pi/3$ gives $a^2 pi$.
$endgroup$
– heropup
Apr 18 '16 at 18:33
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $$r_1(theta) = 3a cos theta$$ is the equation of the circle, and $$r_2theta) = a(1+cos theta)$$ is the equation for the cardioid, then these curves intersect when $r_1(theta) = r_2(theta)$, or $3a cos theta = a(1+cos theta)$, or $$2 cos theta = 1.$$ This gives us the intersection points $$theta = pm frac{pi}{3}.$$ There is another point of intersection at the origin, because the circle is traversed twice for a single traversal of the cardioid. But for our purposes, we can integrate on $theta in [-pi/3, pi/3]$. The area enclosed between $r_1$ and $r_2$ is then given by $$A = int_{theta = -pi/3}^{pi/3} frac{1}{2}left(r_1^2(theta) - r_2^2(theta)right) , dtheta.$$ Note this is not the same as $$int_{theta = -pi/3}^{pi/2} frac{1}{2}left(r_1(theta) - r_2(theta)right)^2 , dtheta.$$ The second integral is incorrect for the area enclosed.
answered Apr 18 '16 at 18:01
heropupheropup
62.9k66099
$endgroup$
$begingroup$
I have used your answer and was left with a^2 pi + a^2 (sqrt(3)/2). Is that correct?
$endgroup$
– user332320
Apr 18 '16 at 18:27
$begingroup$
@user332320 No, it is not correct. $$r_1^2(theta) - r_2^2(theta) = 9a^2 cos^2 theta - a^2 (1+costheta)^2 = a^2 (8 cos^2 theta - 2 cos theta - 1).$$ The antiderivative is $$a^2 (2 sin 2theta - 2 sin theta + 3theta).$$ Evaluating that at $theta = pi/3$ gives $a^2 pi$.
$endgroup$
– heropup
Apr 18 '16 at 18:33
add a comment |
$begingroup$
If $$r_1(theta) = 3a cos theta$$ is the equation of the circle, and $$r_2theta) = a(1+cos theta)$$ is the equation for the cardioid, then these curves intersect when $r_1(theta) = r_2(theta)$, or $3a cos theta = a(1+cos theta)$, or $$2 cos theta = 1.$$ This gives us the intersection points $$theta = pm frac{pi}{3}.$$ There is another point of intersection at the origin, because the circle is traversed twice for a single traversal of the cardioid. But for our purposes, we can integrate on $theta in [-pi/3, pi/3]$. The area enclosed between $r_1$ and $r_2$ is then given by $$A = int_{theta = -pi/3}^{pi/3} frac{1}{2}left(r_1^2(theta) - r_2^2(theta)right) , dtheta.$$ Note this is not the same as $$int_{theta = -pi/3}^{pi/2} frac{1}{2}left(r_1(theta) - r_2(theta)right)^2 , dtheta.$$ The second integral is incorrect for the area enclosed.
answered Apr 18 '16 at 18:01
heropupheropup
62.9k66099
$endgroup$
$begingroup$
I have used your answer and was left with a^2 pi + a^2 (sqrt(3)/2). Is that correct?
$endgroup$
– user332320
Apr 18 '16 at 18:27
$begingroup$
@user332320 No, it is not correct. $$r_1^2(theta) - r_2^2(theta) = 9a^2 cos^2 theta - a^2 (1+costheta)^2 = a^2 (8 cos^2 theta - 2 cos theta - 1).$$ The antiderivative is $$a^2 (2 sin 2theta - 2 sin theta + 3theta).$$ Evaluating that at $theta = pi/3$ gives $a^2 pi$.
$endgroup$
– heropup
Apr 18 '16 at 18:33
add a comment |
$begingroup$
If $$r_1(theta) = 3a cos theta$$ is the equation of the circle, and $$r_2theta) = a(1+cos theta)$$ is the equation for the cardioid, then these curves intersect when $r_1(theta) = r_2(theta)$, or $3a cos theta = a(1+cos theta)$, or $$2 cos theta = 1.$$ This gives us the intersection points $$theta = pm frac{pi}{3}.$$ There is another point of intersection at the origin, because the circle is traversed twice for a single traversal of the cardioid. But for our purposes, we can integrate on $theta in [-pi/3, pi/3]$. The area enclosed between $r_1$ and $r_2$ is then given by $$A = int_{theta = -pi/3}^{pi/3} frac{1}{2}left(r_1^2(theta) - r_2^2(theta)right) , dtheta.$$ Note this is not the same as $$int_{theta = -pi/3}^{pi/2} frac{1}{2}left(r_1(theta) - r_2(theta)right)^2 , dtheta.$$ The second integral is incorrect for the area enclosed.
answered Apr 18 '16 at 18:01
heropupheropup
62.9k66099
$endgroup$
If $$r_1(theta) = 3a cos theta$$ is the equation of the circle, and $$r_2theta) = a(1+cos theta)$$ is the equation for the cardioid, then these curves intersect when $r_1(theta) = r_2(theta)$, or $3a cos theta = a(1+cos theta)$, or $$2 cos theta = 1.$$ This gives us the intersection points $$theta = pm frac{pi}{3}.$$ There is another point of intersection at the origin, because the circle is traversed twice for a single traversal of the cardioid. But for our purposes, we can integrate on $theta in [-pi/3, pi/3]$. The area enclosed between $r_1$ and $r_2$ is then given by $$A = int_{theta = -pi/3}^{pi/3} frac{1}{2}left(r_1^2(theta) - r_2^2(theta)right) , dtheta.$$ Note this is not the same as $$int_{theta = -pi/3}^{pi/2} frac{1}{2}left(r_1(theta) - r_2(theta)right)^2 , dtheta.$$ The second integral is incorrect for the area enclosed.
answered Apr 18 '16 at 18:01
heropupheropup
62.9k66099
edited Apr 18 '16 at 18:07
answered Apr 18 '16 at 18:01
heropupheropup
62.9k66099
answered Apr 18 '16 at 18:01
heropupheropup
62.9k66099
answered Apr 18 '16 at 18:01
heropupheropup
62.9k66099
62.9k66099
$begingroup$
I have used your answer and was left with a^2 pi + a^2 (sqrt(3)/2). Is that correct?
$endgroup$
– user332320
Apr 18 '16 at 18:27
$begingroup$
@user332320 No, it is not correct. $$r_1^2(theta) - r_2^2(theta) = 9a^2 cos^2 theta - a^2 (1+costheta)^2 = a^2 (8 cos^2 theta - 2 cos theta - 1).$$ The antiderivative is $$a^2 (2 sin 2theta - 2 sin theta + 3theta).$$ Evaluating that at $theta = pi/3$ gives $a^2 pi$.
$endgroup$
– heropup
Apr 18 '16 at 18:33
add a comment |
$begingroup$
I have used your answer and was left with a^2 pi + a^2 (sqrt(3)/2). Is that correct?
$endgroup$
– user332320
Apr 18 '16 at 18:27
$begingroup$
@user332320 No, it is not correct. $$r_1^2(theta) - r_2^2(theta) = 9a^2 cos^2 theta - a^2 (1+costheta)^2 = a^2 (8 cos^2 theta - 2 cos theta - 1).$$ The antiderivative is $$a^2 (2 sin 2theta - 2 sin theta + 3theta).$$ Evaluating that at $theta = pi/3$ gives $a^2 pi$.
$endgroup$
– heropup
Apr 18 '16 at 18:33
$begingroup$
I have used your answer and was left with a^2 pi + a^2 (sqrt(3)/2). Is that correct?
$endgroup$
– user332320
Apr 18 '16 at 18:27
$begingroup$
I have used your answer and was left with a^2 pi + a^2 (sqrt(3)/2). Is that correct?
$endgroup$
– user332320
Apr 18 '16 at 18:27
$begingroup$
@user332320 No, it is not correct. $$r_1^2(theta) - r_2^2(theta) = 9a^2 cos^2 theta - a^2 (1+costheta)^2 = a^2 (8 cos^2 theta - 2 cos theta - 1).$$ The antiderivative is $$a^2 (2 sin 2theta - 2 sin theta + 3theta).$$ Evaluating that at $theta = pi/3$ gives $a^2 pi$.
$endgroup$
– heropup
Apr 18 '16 at 18:33
$begingroup$
@user332320 No, it is not correct. $$r_1^2(theta) - r_2^2(theta) = 9a^2 cos^2 theta - a^2 (1+costheta)^2 = a^2 (8 cos^2 theta - 2 cos theta - 1).$$ The antiderivative is $$a^2 (2 sin 2theta - 2 sin theta + 3theta).$$ Evaluating that at $theta = pi/3$ gives $a^2 pi$.
$endgroup$
– heropup
Apr 18 '16 at 18:33
add a comment |
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