[amp_mcq option1=”$$\\frac{2}{3}{{\\text{a}}^2}$$” option2=”$$\\frac{{14}}{3}{{\\text{a}}^2}$$” option3=”$$\\frac{{16}}{3}{{\\text{a}}^2}$$” option4=”$$\\frac{{17}}{3}{{\\text{a}}^2}$$” correct=”option2″]<\/p>\n
The correct answer is $\\boxed{\\frac{16}{3}a^2}$.<\/p>\n
The parabolas $y^2 = 4ax$ and $x^2 = 4ay$ intersect at the points $(\\pm a, \\pm a)$. The area between the parabolas is bounded by the lines $y = x$ and $y = -x$.<\/p>\n
To find the area, we can use the following formula:<\/p>\n
$$\\text{Area} = \\int_{-a}^a \\left( x – y \\right) \\, dy$$<\/p>\n
Substituting in the equations for the parabolas, we get:<\/p>\n
$$\\text{Area} = \\int_{-a}^a \\left( x – \\sqrt{4ax} \\right) \\, dy$$<\/p>\n
We can simplify this integral by using the substitution $u = \\sqrt{4ax}$. This gives us:<\/p>\n
$$\\text{Area} = \\int_{-2a}^2 \\left( x – u \\right) \\, du$$<\/p>\n
We can then evaluate this integral using the following formula:<\/p>\n
$$\\int u^n \\, du = \\frac{u^{n+1}}{n+1} + C$$<\/p>\n
This gives us:<\/p>\n
$$\\text{Area} = \\left[ \\frac{x^2}{2} – \\frac{u^2}{2} \\right]_{-2a}^2 = \\frac{4a^2}{2} – \\frac{4a^2}{2} + \\frac{4a^2}{2} – \\frac{4a^2}{2} = \\boxed{\\frac{16}{3}a^2}$$<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$$\\frac{2}{3}{{\\text{a}}^2}$$” option2=”$$\\frac{{14}}{3}{{\\text{a}}^2}$$” option3=”$$\\frac{{16}}{3}{{\\text{a}}^2}$$” option4=”$$\\frac{{17}}{3}{{\\text{a}}^2}$$” correct=”option2″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20255","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n