Question

Let R be bounded by the curves y = x2 + 2 and y = 2x + 5. What integrals will give the volume of the solid of revolution formed by revolving R around the line x = −3?

Answer 1

Solution:

Given, curve $C:y=x^2+2,$ and line $\ L:y=2x+5$

Around the line: $x=-3$

Outer radius: $x=\sqrt{y-2}+1$

Inner radius: $x=\dfrac{y-5}2+1=\dfrac{y-3}2$

From graph, we can see limits of y are 2 to 11.

So, a=2, b=11.

Now, area, $A(y)=\pi[(\sqrt{y-2}+1)^2-(\dfrac{y-3}2)^2]$

$\Rightarrow A(y)=\pi[(y-2+1+2\sqrt{y-2})-(\dfrac{y^2+9-6y}4)]$

$\Rightarrow A(y)=\pi[y-1+2\sqrt{y-2}-\dfrac{y^2}4 -\dfrac94+\dfrac{3y}2]$

$\Rightarrow A(y)=\pi(2\sqrt{y-2}-\dfrac{y^2}4 -\dfrac{13}4+\dfrac{5y}2)$

Now, Volume, $V=\int_{2}^{11}[\pi(2\sqrt{y-2}-\dfrac{y^2}4 -\dfrac{13}4+\dfrac{5y}2)]dy$ $\Rightarrow V=\pi[(\dfrac{4(y-2)^{3/2}}{3}-\dfrac{y^3}{12} -\dfrac{13y}4+\dfrac{5y^2}4)]_{2}^{11}$

$\Rightarrow V=\pi[(\dfrac{4(11-2)^{3/2}}{3}-\dfrac{{11}^3}{12} -\dfrac{13(11)}4+\dfrac{5(11)^2}4)-(\dfrac{4(2-2)^{3/2}}{3}-\dfrac{2^3}{12} -\dfrac{13(2)}4+\dfrac{5(2)^2}4)]$

$\Rightarrow V=\pi[(36-\dfrac{1331}{12} -\dfrac{143}4+\dfrac{605}4-(0-\dfrac{2}{3} -\dfrac{13}2+5)]$

$\Rightarrow V=\pi[36-\dfrac{1331}{12} +\dfrac{231}2+\dfrac{2}{3} +\dfrac{13}2-5]$

$\Rightarrow V=\pi[36-\dfrac{1331}{12} +\dfrac{231}2+\dfrac{2}{3} +\dfrac{13}2-5]$

$\Rightarrow V=42.75\pi\ \text{units}^3$