Intergrals - Sums and differences of areas (1 Viewer)

V_L · Mar 25, 2017

Hi,

Im not sure of the way you do these questions if someone could please explain,

Find the area bounded by the line y = 1 and the curve y = x 2 .

Find the area bounded by the curve y = 9 - x 2 and the line y = 5.

1729 · Mar 25, 2017

V_L said:
Hi,

Im not sure of the way you do these questions if someone could please explain,

Find the area bounded by the line y = 1 and the curve y = x 2 .

Find the area bounded by the curve y = 9 - x 2 and the line y = 5.

$\noindent For the first one, the area $A$ is simply $A = \int_{-1}^1 \mathrm{d}x - \int_{-1}^1x^2 \mathrm{d}x$ since the points of intersection are $(-1, 1)$ and $(1, 1)$. This can be written as $A = \int_{-1}^{1}(1-x^2)\mathrm{d}x$ by combining the integrals. Now by symmetry, this becomes $A = 2 \times \int_0^1 (1-x^2)\mathrm{d}x = 2\left[x - \frac{x^3}{3}\right]^1_0 = 2\left[1 - \frac{1}{3} - 0\right] = 1\frac{1}{3}$ square units. \\\\ Here, we found the area under the line $y = 1$ and subtracted the area under the curve $y = x^2$

1729 · Mar 25, 2017

V_L said:
Hi,

Im not sure of the way you do these questions if someone could please explain,

Find the area bounded by the line y = 1 and the curve y = x 2 .

Find the area bounded by the curve y = 9 - x 2 and the line y = 5.

$$\noindent There are multiple ways to do the second question, one being taking the difference of areas by integrating with respect to $x$. \\\\ Alternatively, we can use the fact that if $y=9-x^2$ then $x = \pm \sqrt{9-y}$. By symmetry, the required area $A = 2 \times \int_5^9 (\sqrt{9-y})\mathrm{d}y$, as in the first quadrant we use $x = \sqrt{9-y}$. That is, $A = 2\left[-\frac{2(9-y)^{\frac{3}{2}}}{3}\right]^9_5 = 10\frac{2}{3}$ $u^2$

$\noindent Other method is to use $A = \int_{-2}^2 (9-x^2)\mathrm{d}x - \int_{-2}^2 5\mathrm{d}x = \int_{-2}^2 (9-x^2 - 5)\mathrm{d}x = \int_{-2}^2(4-x^2)\mathrm{d}x.$ By symmetry this becomes $A = 2 \times \int_0^2 (4-x^2)\mathrm{d}x$. Note that the intersection points are at $x = -2$ and $x=2$. Here, we found the area under the curve $y = 9 - x^2$ and subtracted the area under the line $y = 5$

V_L · Mar 26, 2017

Thank you so much!!

Intergrals - Sums and differences of areas (1 Viewer)

V_L

New Member

1729

Active Member

1729

Active Member

V_L

New Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)