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Previous Year Question (PYQs)



The volume of the solid generated by revolving the region between the y-axis and the curve $x=2\sqrt{y}, \ 0\le y\le4$ about the y-axis is





Solution

Using washer method: $V=\pi\int_0^4 (2\sqrt{y})^2 dy$ $=\pi\int_0^4 4y,dy$ $=4\pi\left[\frac{y^2}{2}\right]_0^4$ $=4\pi \times 8$ $=32\pi$


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