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So, I am calculating the volume by revolving the region between the curves $$2x+3y=6\quad\text{and}\quad (y-1)^2=4-x$$

about

$i)$ $x=-5$

$ii)$ $y=0$

The problem is though I can't find the intersection points. I get an unbounded integral if I try to do that and I don't think that's that right approach...

Can some give me a suggestion as to how I would go about doing this problem? Thanks!

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    Here is a link to a graph of your region. Are you sure about (ii) $x=0$ ? That axis goes through the region. Should it be $y=0$ instead. https://www.desmos.com/calculator/d4gapyydlw – 2017-02-18
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    Oops yeah y=0. One moment while I edit it. – 2017-02-18
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    For the $x=-5$ axis you should use the annulus method method and for the $y=0$ axis you should use the cylindrical shell method. For both you will integrate with respect to $y$ between $y=0$ and $y=7/2$. – 2017-02-18
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    To get the intersection of the two curves it is easier to solve each of the two equations for $x$ in terms of $y$ and set the two $y$ expressions equal to each other. This will give a quadratic equation in $y$ which can be easily factored so that both $y$ coordinates of the intersection points can be found. – 2017-02-18
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    Yep I got the intersection points already. I'm just going to do the integral now. Thanks a lot for this :) . – 2017-02-18
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    You are welcome. – 2017-02-18

1 Answers 1

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The region is given below.

enter image description here

The intersection occur at the points where $y=0$ and $y=\frac{7}{2}$. Let us write $$g(y)=4-(y-1)^2=-y^2+2y+3$$ and $$f(y)=\frac{6-3y}{2}.$$ If the region $R$ is revolved about the line $x=-5$, then the volume of the solid generated (by using the washer method) is given by $$V=\int_{0}^{\frac{7}{2}}\pi\bigg(\big[g(y)-(-5)\big]^2-\big[f(y)-(-5)\big]^2\bigg)dy$$ and if the region $R$ is revolved about the line $y=0$ then the volume of the solid generated (by using the shell method) is given by $$V=\int_{0}^{\frac{7}{2}}2\pi y\cdot\big[g(y)-f(y)\big]dy$$ Can you proceed with the computations?

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    I almost got the exact same thing but for the second integral, why is the radius just $y$? I thought it would be $\frac{6-3y}{2}$. Everything else is the same as yours however so that's a good sign. – 2017-02-18
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    Are you referring to the shell method? – 2017-02-18
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    Yes, I am referring to the shell method. – 2017-02-18
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    The distance of the sliced element, which is a horizontal strip, is $y$ because the region is to be revolved about the line $y=0$ or known as the $x$-axis. Thus, the circumference of the circle generated is given by $2\pi y$. The area of the strip is $[g(y)-f(y)]dy$. – 2017-02-18
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    Ohh yes yes. I forgot my my variables were in terms of y and not x. That makes sense. Thank you. – 2017-02-18
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    BTW, I did not yet compute the two integrals. Hope you continue. Thanks. – 2017-02-18
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    I don't need to. The question only asked me to set it up so I am done at this point. – 2017-02-18
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    How do you know if the axis of revolution is $x$ or $y$ ?? – 2017-02-27
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    @Yves Daoust I don't get what you mean. The OP wanted only that axis of revolution is the line $x=-5$ and the line $y=0$ which is the $x$-axis. – 2017-02-27
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    @ΘΣΦGenSan: sorry, my bad, ignore my comment. – 2017-02-27
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    @Yves Daoust No worries Sir. – 2017-02-27