Finding volume of solid (using shell method) obtained by revolving the region formed by $x=2\sqrt{y}\space ,y=-x\space, y=4$ about $x$ axis is
What i try: first we will solve $y=-x $ and $y=4$, we get $x=-4$
And we will solve $x=2\sqrt{y}$ and $y=4$, we get $x=4$
So volume of solid formed by rotating the region formed by $x=2\sqrt{y},y=-x$ and $y=2$ using shell method is
$$V=2\pi\int^{0}_{-4}\bigg[4-(-x)\bigg]xdx+2\pi\int^{4}_{0}\bigg[4-\frac{x^2}{4}\bigg]xdx$$
Can anyone please tell me is i am right or wrong? If wrong Then please tell me right answer
No, it is not correct. If you want to compute this volume through the shell method, what you should compute is:$$2\pi\int_0^4y\left(2\sqrt y+y\right)\,\mathrm dy=\frac{1\,408}{15}\pi.$$Or you can use the disk method:$$\pi\int_{-4}^04^2-(-x)^2\,\mathrm dx+\pi\int_0^44^2-\left(\frac{x^2}4\right)^2\,\mathrm dx=\frac{1\,408}{15}\pi.$$