Pull and Twist
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The question is...
So of course I rearrange my ellipse formula to get $$y=\pm\sqrt{1-\frac{{x}^{2}}{4}}$$
Then I calculate my radius as $$x-(-2)=r \implies x+2=r$$
I know the formula for finding volume with the Shell Method is $$V=\int_{a}^{b}2\pi x\cdot f(x) \,dx$$
To make it easier I decide to just find the volume of the top half and times my equation by two. Having graphed the equation I know I need to integrate between $$-2\le x\le2$$
So I then have...
$$2\int_{-2}^{2} 2\pi(x+2)(\sqrt{1-\frac{{x}^{2}}{4}})\,dx$$
$$4\pi\left\{\int_{-2}^{2} x(\sqrt{1-\frac{{x}^{2}}{4}})\,dx + \int_{-2}^{2} 2(\sqrt{1-\frac{{x}^{2}}{4}})\,dx\right\}$$
$$4\pi\left\{\int_{-2}^{2} x(\sqrt{1-\frac{{x}^{2}}{4}})\,dx + 2\int_{-2}^{2}(\sqrt{1-\frac{{x}^{2}}{4}})\,dx\right\}$$
Using $$u=1-\frac{{x}^{4}}{4}$$ substitution for the first integral I get...
$$4\pi\left\{-\frac{4}{3}\left[{\left(1-\frac{{x}^{2}}{4}\right)}^{\frac{3}{2}}\right]_{-2}^{2} + 2\int_{-2}^{2}(\sqrt{1-\frac{{x}^{2}}{4}})\,dx\right\}$$
Am I taking the right approach?? If so... how do I integrate the second integral w/o using a table of integrals?? Would this problem be easier utilizing the disk/washer method?
I need me some help. Thanks.
Take-Home Problem said:Find the volume of the solid obtained by rotating the ellipse $${x}^{2}+4{y}^{2}=4$$ about $$x=-2$$ using the Shell Method. *Table of Integrals is not allowed for integration.
So of course I rearrange my ellipse formula to get $$y=\pm\sqrt{1-\frac{{x}^{2}}{4}}$$
Then I calculate my radius as $$x-(-2)=r \implies x+2=r$$
I know the formula for finding volume with the Shell Method is $$V=\int_{a}^{b}2\pi x\cdot f(x) \,dx$$
To make it easier I decide to just find the volume of the top half and times my equation by two. Having graphed the equation I know I need to integrate between $$-2\le x\le2$$
So I then have...
$$2\int_{-2}^{2} 2\pi(x+2)(\sqrt{1-\frac{{x}^{2}}{4}})\,dx$$
$$4\pi\left\{\int_{-2}^{2} x(\sqrt{1-\frac{{x}^{2}}{4}})\,dx + \int_{-2}^{2} 2(\sqrt{1-\frac{{x}^{2}}{4}})\,dx\right\}$$
$$4\pi\left\{\int_{-2}^{2} x(\sqrt{1-\frac{{x}^{2}}{4}})\,dx + 2\int_{-2}^{2}(\sqrt{1-\frac{{x}^{2}}{4}})\,dx\right\}$$
Using $$u=1-\frac{{x}^{4}}{4}$$ substitution for the first integral I get...
$$4\pi\left\{-\frac{4}{3}\left[{\left(1-\frac{{x}^{2}}{4}\right)}^{\frac{3}{2}}\right]_{-2}^{2} + 2\int_{-2}^{2}(\sqrt{1-\frac{{x}^{2}}{4}})\,dx\right\}$$
Am I taking the right approach?? If so... how do I integrate the second integral w/o using a table of integrals?? Would this problem be easier utilizing the disk/washer method?
I need me some help. Thanks.