MHB 15.2.78 But it asks for a double integral

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The discussion focuses on calculating the area of the region bounded by the curves y=4+4sin(x) and y=4-4sin(x) over the interval [0, π] using a double integral. Participants note that while the problem could be solved more simply with a single integral, it specifically requests a double integral format. The suggested double integral is expressed as ∫₀^π ∫₄⁻⁴sin(x)⁴⁺⁴sin(x) dy dx. This approach maintains the formal requirement of a double integral while effectively simplifying to a single integral calculation. The discussion highlights the nuances of interpreting integral types in mathematical problems.
karush
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Use double integral to compute the area of the region
bounded by $y=4+4\sin{x}$ and $y=4-4\sin{x}$
on the interval $\left[0,\pi\right]$

View attachment 7253

ok it looks easier to do this in one $\int$ but it asks for a double $\int\int$ so ?
 

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Re: 15.2.78 but it asks for a double integral

karush said:
Use double integral to compute the area of the region
bounded by $y=4+4\sin{x}$ and $y=4-4\sin{x}$
on the interval $\left[0,\pi\right]$
ok it looks easier to do this in one $\int$ but it asks for a double $\int\int$ so ?
$$\int_0^\pi\int_{4-4\sin x}^{4+4\sin x}dy\,dx$$ Okay, it's just a single integral in disguise, but formally it's a double integral.