I found out the area of a heart but I am not sure whether it is correct.
Can someone please check whether what I have done makes any sense at all.
The equation of a heart is shown below:
I first determined the jacobian(shown below):
Then I integrated the determinant of the jacobian( shown below):
and I got: $9π$/$4$ $units^3$
I sorry about the size of some of the formulas.
$\endgroup$ 121 Answer
$\begingroup$You can save yourself some of the tedious algebraic and trigonometric manipulation by using Green’s Theorem. Since $\mathrm d(y\;\mathrm dx)=\mathrm dx\wedge\mathrm dy$, the integral $\int_Cy\;\mathrm dx$ gives the area of the region bounded by the closed curve $C$. In this case, we have $$\begin{align} \int_Cy\;\mathrm dx &= \int_0^{2\pi}(3\cos t-2\cos{2t}-\frac12\cos{3t})\cdot6\sin^2t\cos t\;dt \\ &= \frac18\int_0^{2\pi}21-3\cos{2t}-12\cos{3t}-21\cos{4t}+12\cos{5t}+3\cos{6t}\;dt \\ &= \frac{21\pi}4. \end{align}$$ Note that you can add the differential $\mathrm df$ of any differentiable function defined on the region to the integrand in order to simplify it without changing the value of the integral. A common choice is to integrate $\frac12(x\;\mathrm dy-y\;\mathrm dx)$, but that doesn’t really help here.
$\endgroup$
