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Find the total surface area of a hemispherical bowl of thickness 'd' and internal radius 'r'.
$$4\pi r^{2} + 4\pi rd + 3d^{2}$$
$$\pi(4r^{2} + 3rd + d^{2}$$)
$$\pi(4r^{2} + 6rd + 3d^{2})$$
$$4πr^{2} + 6πrd + 3d^{2}$$
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